Washer method confusion












2












$begingroup$


Compute volume of solid created by revolving the area bounded by $y=sqrt x$, $y=0$ $x=1$, $x=4$, around the y axis.
I understand that I can find the volume by integration of the $A(y)$ from $0$ to $2$ since these are the $y$ values of the intercepts of $x=1$ and $x=4$.



My understanding is that I would compute the area $(A(y) $by:
$pi$ $(R^2)-r^2)$ where $R$ is the outer radius and $r$ is the inner radius. Then I integrate this as $int_0^2 A(y)dy$.



So, I compute the outer radius $R$ by calculating the $x$ distance from the right most boundary, in this case $x=4$ and the $y$ axis. Thus $R=4$. Is this the way to calculate $R$ even if part of the $R$ $x$ distance isn't inside the boundary?? Ugh,confused on the definition here.



The inner radius is the $x$ distance from the function to the axis of rotation (the $y$ axis). So, $r=y^2$.



But, it I would think that this x distance is ONLY within the region of $x=1$ and $x=4$. So, since I should always calculate from right to left, wouldn't this radius be $y^2-1$?? But seems like I am using the $R$ by NOT considering the area in the bounded region and I AM using the $r$ by considering the bonded region.



I cannot seem to understand this washer problem even though I can work others and I can see that I don't fully understand the definition of how to computer the inner and outer radii.



Can someone clarify? Bottom line is that I can't seem to be able to compute the inner and outer radii of this problem correctly.










share|cite|improve this question









$endgroup$

















    2












    $begingroup$


    Compute volume of solid created by revolving the area bounded by $y=sqrt x$, $y=0$ $x=1$, $x=4$, around the y axis.
    I understand that I can find the volume by integration of the $A(y)$ from $0$ to $2$ since these are the $y$ values of the intercepts of $x=1$ and $x=4$.



    My understanding is that I would compute the area $(A(y) $by:
    $pi$ $(R^2)-r^2)$ where $R$ is the outer radius and $r$ is the inner radius. Then I integrate this as $int_0^2 A(y)dy$.



    So, I compute the outer radius $R$ by calculating the $x$ distance from the right most boundary, in this case $x=4$ and the $y$ axis. Thus $R=4$. Is this the way to calculate $R$ even if part of the $R$ $x$ distance isn't inside the boundary?? Ugh,confused on the definition here.



    The inner radius is the $x$ distance from the function to the axis of rotation (the $y$ axis). So, $r=y^2$.



    But, it I would think that this x distance is ONLY within the region of $x=1$ and $x=4$. So, since I should always calculate from right to left, wouldn't this radius be $y^2-1$?? But seems like I am using the $R$ by NOT considering the area in the bounded region and I AM using the $r$ by considering the bonded region.



    I cannot seem to understand this washer problem even though I can work others and I can see that I don't fully understand the definition of how to computer the inner and outer radii.



    Can someone clarify? Bottom line is that I can't seem to be able to compute the inner and outer radii of this problem correctly.










    share|cite|improve this question









    $endgroup$















      2












      2








      2


      0



      $begingroup$


      Compute volume of solid created by revolving the area bounded by $y=sqrt x$, $y=0$ $x=1$, $x=4$, around the y axis.
      I understand that I can find the volume by integration of the $A(y)$ from $0$ to $2$ since these are the $y$ values of the intercepts of $x=1$ and $x=4$.



      My understanding is that I would compute the area $(A(y) $by:
      $pi$ $(R^2)-r^2)$ where $R$ is the outer radius and $r$ is the inner radius. Then I integrate this as $int_0^2 A(y)dy$.



      So, I compute the outer radius $R$ by calculating the $x$ distance from the right most boundary, in this case $x=4$ and the $y$ axis. Thus $R=4$. Is this the way to calculate $R$ even if part of the $R$ $x$ distance isn't inside the boundary?? Ugh,confused on the definition here.



      The inner radius is the $x$ distance from the function to the axis of rotation (the $y$ axis). So, $r=y^2$.



      But, it I would think that this x distance is ONLY within the region of $x=1$ and $x=4$. So, since I should always calculate from right to left, wouldn't this radius be $y^2-1$?? But seems like I am using the $R$ by NOT considering the area in the bounded region and I AM using the $r$ by considering the bonded region.



      I cannot seem to understand this washer problem even though I can work others and I can see that I don't fully understand the definition of how to computer the inner and outer radii.



      Can someone clarify? Bottom line is that I can't seem to be able to compute the inner and outer radii of this problem correctly.










      share|cite|improve this question









      $endgroup$




      Compute volume of solid created by revolving the area bounded by $y=sqrt x$, $y=0$ $x=1$, $x=4$, around the y axis.
      I understand that I can find the volume by integration of the $A(y)$ from $0$ to $2$ since these are the $y$ values of the intercepts of $x=1$ and $x=4$.



      My understanding is that I would compute the area $(A(y) $by:
      $pi$ $(R^2)-r^2)$ where $R$ is the outer radius and $r$ is the inner radius. Then I integrate this as $int_0^2 A(y)dy$.



      So, I compute the outer radius $R$ by calculating the $x$ distance from the right most boundary, in this case $x=4$ and the $y$ axis. Thus $R=4$. Is this the way to calculate $R$ even if part of the $R$ $x$ distance isn't inside the boundary?? Ugh,confused on the definition here.



      The inner radius is the $x$ distance from the function to the axis of rotation (the $y$ axis). So, $r=y^2$.



      But, it I would think that this x distance is ONLY within the region of $x=1$ and $x=4$. So, since I should always calculate from right to left, wouldn't this radius be $y^2-1$?? But seems like I am using the $R$ by NOT considering the area in the bounded region and I AM using the $r$ by considering the bonded region.



      I cannot seem to understand this washer problem even though I can work others and I can see that I don't fully understand the definition of how to computer the inner and outer radii.



      Can someone clarify? Bottom line is that I can't seem to be able to compute the inner and outer radii of this problem correctly.







      calculus






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      share|cite|improve this question










      asked Dec 29 '18 at 0:30









      user163862user163862

      88021017




      88021017






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          Cross section



          enter image description here





          Shell method



          enter image description here



          $$V = intlimits_{r=1}^4 2 pi r sqrt{r} dr = frac{124 pi }{5}$$





          Washer method



          enter image description here



          $$V = intlimits_{z=0}^1 pi (4^2 - 1^2) dz + intlimits_{z=1}^2 pi (4^2 - z^2) dz = frac{124 pi }{5}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The calculation at the bottom looks like some kind of unholy hybrid between washers and shells. Which is it supposed to be?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:42












          • $begingroup$
            These are shells.
            $endgroup$
            – David G. Stork
            Dec 29 '18 at 1:43






          • 1




            $begingroup$
            Then it should be $2pi r$ rather than $pi(r^2-1^2)$, namely the circumrefence of the shell at radius $r$.
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:44










          • $begingroup$
            Yep (+1)... thanks..
            $endgroup$
            – David G. Stork
            Dec 29 '18 at 1:46












          • $begingroup$
            Then I only take issue with your value for the definite integral. It looks like you may have forgotten to take an antiderivative ... or something worse. (The volume of the entire circumscribed cylinder is only $32pi$).
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:50





















          2












          $begingroup$

          First, draw a diagram. This step is mandatory and to be skipped only if you have a clear mental picture of how it will look without drawing it.



          diagram of the are to be rotated



          Since we're rotating about the $y$-axis the washers are horizontal sections through the shaded areas -- and the radius is just the $x$ coordinate of each point in the figure. We can see on the diagram that the range of relevant $y$s is $[0,2]$, and the inner and outer radiuses will be:



          $$ {rm inner}(y) = begin{cases} 1 & text{for }0le y le 1 \ y^2 & text{for } 1 le y le 2 end{cases} qquad qquad {rm outer}(y) = 4 $$



          Can you take it from here?





          By the way, the shape of the area suggests it would be much easier to compute with the shell method. You wouldn't need any piecewise defined functions and could do everything in one integral.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I did draw an image similar to yours. I got similar values for inner and outer radii and could NOT get the text answer which is 124pi/5.
            $endgroup$
            – user163862
            Dec 29 '18 at 2:16










          • $begingroup$
            Can you assist me in understanding why there are 2 inner radii?
            $endgroup$
            – user163862
            Dec 29 '18 at 2:30










          • $begingroup$
            @user163862: The left side of the shaded area consists of a straight line from $(1,0)$ to $(1,1)$, and the part of the parabola between $(1,1)$ and $(4,2)$. The sharp bend in this inside edge corresponds to the switch between expressions.
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 3:28





















          0












          $begingroup$

          The inner radius would be $1$ for $0leq y<1$ and $y^2$ for $1leq yleq2$.



          The outer radius would be 4 for $0leq yleq2$



          Then, the set up for the volume would be
          $$V=piint_{0}^{1}(4^2-1^2)~dy+piint_{1}^{2}(4^2-y^4)~dy=frac{124pi}{5}$$
          enter image description here






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How on earth do you get those radiuses?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 0:43










          • $begingroup$
            I used the graph of $x=y^2$. Since the area is bounded between x=4 and x=1, I think the outer radius should be $4-y^2$, and the inner radius should be $1-y^2$. Did I do anything wrong?
            $endgroup$
            – Larry
            Dec 29 '18 at 0:49










          • $begingroup$
            I'm still quite baffled where you're measuring those radii. Are you imagining we're you're rotating about the line $x=4$? No, that doesn't make sense either. Perhaps you could draw a diagram to show more precisely where you get those distances from?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 0:58










          • $begingroup$
            I think I made a mistake, I'll edit my answer.
            $endgroup$
            – Larry
            Dec 29 '18 at 1:01






          • 1




            $begingroup$
            x @Larry: Um ... I can see what you're doing now, but it is quite wrong. "Radius" means distance from the axis of revolution, and the axis of revolution is the y-axis, not the red parabola! How would you rotate something about a curved axis anyway?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:23













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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Cross section



          enter image description here





          Shell method



          enter image description here



          $$V = intlimits_{r=1}^4 2 pi r sqrt{r} dr = frac{124 pi }{5}$$





          Washer method



          enter image description here



          $$V = intlimits_{z=0}^1 pi (4^2 - 1^2) dz + intlimits_{z=1}^2 pi (4^2 - z^2) dz = frac{124 pi }{5}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The calculation at the bottom looks like some kind of unholy hybrid between washers and shells. Which is it supposed to be?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:42












          • $begingroup$
            These are shells.
            $endgroup$
            – David G. Stork
            Dec 29 '18 at 1:43






          • 1




            $begingroup$
            Then it should be $2pi r$ rather than $pi(r^2-1^2)$, namely the circumrefence of the shell at radius $r$.
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:44










          • $begingroup$
            Yep (+1)... thanks..
            $endgroup$
            – David G. Stork
            Dec 29 '18 at 1:46












          • $begingroup$
            Then I only take issue with your value for the definite integral. It looks like you may have forgotten to take an antiderivative ... or something worse. (The volume of the entire circumscribed cylinder is only $32pi$).
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:50


















          1












          $begingroup$

          Cross section



          enter image description here





          Shell method



          enter image description here



          $$V = intlimits_{r=1}^4 2 pi r sqrt{r} dr = frac{124 pi }{5}$$





          Washer method



          enter image description here



          $$V = intlimits_{z=0}^1 pi (4^2 - 1^2) dz + intlimits_{z=1}^2 pi (4^2 - z^2) dz = frac{124 pi }{5}$$






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            The calculation at the bottom looks like some kind of unholy hybrid between washers and shells. Which is it supposed to be?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:42












          • $begingroup$
            These are shells.
            $endgroup$
            – David G. Stork
            Dec 29 '18 at 1:43






          • 1




            $begingroup$
            Then it should be $2pi r$ rather than $pi(r^2-1^2)$, namely the circumrefence of the shell at radius $r$.
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:44










          • $begingroup$
            Yep (+1)... thanks..
            $endgroup$
            – David G. Stork
            Dec 29 '18 at 1:46












          • $begingroup$
            Then I only take issue with your value for the definite integral. It looks like you may have forgotten to take an antiderivative ... or something worse. (The volume of the entire circumscribed cylinder is only $32pi$).
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:50
















          1












          1








          1





          $begingroup$

          Cross section



          enter image description here





          Shell method



          enter image description here



          $$V = intlimits_{r=1}^4 2 pi r sqrt{r} dr = frac{124 pi }{5}$$





          Washer method



          enter image description here



          $$V = intlimits_{z=0}^1 pi (4^2 - 1^2) dz + intlimits_{z=1}^2 pi (4^2 - z^2) dz = frac{124 pi }{5}$$






          share|cite|improve this answer











          $endgroup$



          Cross section



          enter image description here





          Shell method



          enter image description here



          $$V = intlimits_{r=1}^4 2 pi r sqrt{r} dr = frac{124 pi }{5}$$





          Washer method



          enter image description here



          $$V = intlimits_{z=0}^1 pi (4^2 - 1^2) dz + intlimits_{z=1}^2 pi (4^2 - z^2) dz = frac{124 pi }{5}$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 29 '18 at 2:17

























          answered Dec 29 '18 at 1:13









          David G. StorkDavid G. Stork

          11k41432




          11k41432












          • $begingroup$
            The calculation at the bottom looks like some kind of unholy hybrid between washers and shells. Which is it supposed to be?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:42












          • $begingroup$
            These are shells.
            $endgroup$
            – David G. Stork
            Dec 29 '18 at 1:43






          • 1




            $begingroup$
            Then it should be $2pi r$ rather than $pi(r^2-1^2)$, namely the circumrefence of the shell at radius $r$.
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:44










          • $begingroup$
            Yep (+1)... thanks..
            $endgroup$
            – David G. Stork
            Dec 29 '18 at 1:46












          • $begingroup$
            Then I only take issue with your value for the definite integral. It looks like you may have forgotten to take an antiderivative ... or something worse. (The volume of the entire circumscribed cylinder is only $32pi$).
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:50




















          • $begingroup$
            The calculation at the bottom looks like some kind of unholy hybrid between washers and shells. Which is it supposed to be?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:42












          • $begingroup$
            These are shells.
            $endgroup$
            – David G. Stork
            Dec 29 '18 at 1:43






          • 1




            $begingroup$
            Then it should be $2pi r$ rather than $pi(r^2-1^2)$, namely the circumrefence of the shell at radius $r$.
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:44










          • $begingroup$
            Yep (+1)... thanks..
            $endgroup$
            – David G. Stork
            Dec 29 '18 at 1:46












          • $begingroup$
            Then I only take issue with your value for the definite integral. It looks like you may have forgotten to take an antiderivative ... or something worse. (The volume of the entire circumscribed cylinder is only $32pi$).
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:50


















          $begingroup$
          The calculation at the bottom looks like some kind of unholy hybrid between washers and shells. Which is it supposed to be?
          $endgroup$
          – Henning Makholm
          Dec 29 '18 at 1:42






          $begingroup$
          The calculation at the bottom looks like some kind of unholy hybrid between washers and shells. Which is it supposed to be?
          $endgroup$
          – Henning Makholm
          Dec 29 '18 at 1:42














          $begingroup$
          These are shells.
          $endgroup$
          – David G. Stork
          Dec 29 '18 at 1:43




          $begingroup$
          These are shells.
          $endgroup$
          – David G. Stork
          Dec 29 '18 at 1:43




          1




          1




          $begingroup$
          Then it should be $2pi r$ rather than $pi(r^2-1^2)$, namely the circumrefence of the shell at radius $r$.
          $endgroup$
          – Henning Makholm
          Dec 29 '18 at 1:44




          $begingroup$
          Then it should be $2pi r$ rather than $pi(r^2-1^2)$, namely the circumrefence of the shell at radius $r$.
          $endgroup$
          – Henning Makholm
          Dec 29 '18 at 1:44












          $begingroup$
          Yep (+1)... thanks..
          $endgroup$
          – David G. Stork
          Dec 29 '18 at 1:46






          $begingroup$
          Yep (+1)... thanks..
          $endgroup$
          – David G. Stork
          Dec 29 '18 at 1:46














          $begingroup$
          Then I only take issue with your value for the definite integral. It looks like you may have forgotten to take an antiderivative ... or something worse. (The volume of the entire circumscribed cylinder is only $32pi$).
          $endgroup$
          – Henning Makholm
          Dec 29 '18 at 1:50






          $begingroup$
          Then I only take issue with your value for the definite integral. It looks like you may have forgotten to take an antiderivative ... or something worse. (The volume of the entire circumscribed cylinder is only $32pi$).
          $endgroup$
          – Henning Makholm
          Dec 29 '18 at 1:50













          2












          $begingroup$

          First, draw a diagram. This step is mandatory and to be skipped only if you have a clear mental picture of how it will look without drawing it.



          diagram of the are to be rotated



          Since we're rotating about the $y$-axis the washers are horizontal sections through the shaded areas -- and the radius is just the $x$ coordinate of each point in the figure. We can see on the diagram that the range of relevant $y$s is $[0,2]$, and the inner and outer radiuses will be:



          $$ {rm inner}(y) = begin{cases} 1 & text{for }0le y le 1 \ y^2 & text{for } 1 le y le 2 end{cases} qquad qquad {rm outer}(y) = 4 $$



          Can you take it from here?





          By the way, the shape of the area suggests it would be much easier to compute with the shell method. You wouldn't need any piecewise defined functions and could do everything in one integral.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I did draw an image similar to yours. I got similar values for inner and outer radii and could NOT get the text answer which is 124pi/5.
            $endgroup$
            – user163862
            Dec 29 '18 at 2:16










          • $begingroup$
            Can you assist me in understanding why there are 2 inner radii?
            $endgroup$
            – user163862
            Dec 29 '18 at 2:30










          • $begingroup$
            @user163862: The left side of the shaded area consists of a straight line from $(1,0)$ to $(1,1)$, and the part of the parabola between $(1,1)$ and $(4,2)$. The sharp bend in this inside edge corresponds to the switch between expressions.
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 3:28


















          2












          $begingroup$

          First, draw a diagram. This step is mandatory and to be skipped only if you have a clear mental picture of how it will look without drawing it.



          diagram of the are to be rotated



          Since we're rotating about the $y$-axis the washers are horizontal sections through the shaded areas -- and the radius is just the $x$ coordinate of each point in the figure. We can see on the diagram that the range of relevant $y$s is $[0,2]$, and the inner and outer radiuses will be:



          $$ {rm inner}(y) = begin{cases} 1 & text{for }0le y le 1 \ y^2 & text{for } 1 le y le 2 end{cases} qquad qquad {rm outer}(y) = 4 $$



          Can you take it from here?





          By the way, the shape of the area suggests it would be much easier to compute with the shell method. You wouldn't need any piecewise defined functions and could do everything in one integral.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            I did draw an image similar to yours. I got similar values for inner and outer radii and could NOT get the text answer which is 124pi/5.
            $endgroup$
            – user163862
            Dec 29 '18 at 2:16










          • $begingroup$
            Can you assist me in understanding why there are 2 inner radii?
            $endgroup$
            – user163862
            Dec 29 '18 at 2:30










          • $begingroup$
            @user163862: The left side of the shaded area consists of a straight line from $(1,0)$ to $(1,1)$, and the part of the parabola between $(1,1)$ and $(4,2)$. The sharp bend in this inside edge corresponds to the switch between expressions.
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 3:28
















          2












          2








          2





          $begingroup$

          First, draw a diagram. This step is mandatory and to be skipped only if you have a clear mental picture of how it will look without drawing it.



          diagram of the are to be rotated



          Since we're rotating about the $y$-axis the washers are horizontal sections through the shaded areas -- and the radius is just the $x$ coordinate of each point in the figure. We can see on the diagram that the range of relevant $y$s is $[0,2]$, and the inner and outer radiuses will be:



          $$ {rm inner}(y) = begin{cases} 1 & text{for }0le y le 1 \ y^2 & text{for } 1 le y le 2 end{cases} qquad qquad {rm outer}(y) = 4 $$



          Can you take it from here?





          By the way, the shape of the area suggests it would be much easier to compute with the shell method. You wouldn't need any piecewise defined functions and could do everything in one integral.






          share|cite|improve this answer











          $endgroup$



          First, draw a diagram. This step is mandatory and to be skipped only if you have a clear mental picture of how it will look without drawing it.



          diagram of the are to be rotated



          Since we're rotating about the $y$-axis the washers are horizontal sections through the shaded areas -- and the radius is just the $x$ coordinate of each point in the figure. We can see on the diagram that the range of relevant $y$s is $[0,2]$, and the inner and outer radiuses will be:



          $$ {rm inner}(y) = begin{cases} 1 & text{for }0le y le 1 \ y^2 & text{for } 1 le y le 2 end{cases} qquad qquad {rm outer}(y) = 4 $$



          Can you take it from here?





          By the way, the shape of the area suggests it would be much easier to compute with the shell method. You wouldn't need any piecewise defined functions and could do everything in one integral.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 29 '18 at 0:59

























          answered Dec 29 '18 at 0:54









          Henning MakholmHenning Makholm

          241k17308546




          241k17308546












          • $begingroup$
            I did draw an image similar to yours. I got similar values for inner and outer radii and could NOT get the text answer which is 124pi/5.
            $endgroup$
            – user163862
            Dec 29 '18 at 2:16










          • $begingroup$
            Can you assist me in understanding why there are 2 inner radii?
            $endgroup$
            – user163862
            Dec 29 '18 at 2:30










          • $begingroup$
            @user163862: The left side of the shaded area consists of a straight line from $(1,0)$ to $(1,1)$, and the part of the parabola between $(1,1)$ and $(4,2)$. The sharp bend in this inside edge corresponds to the switch between expressions.
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 3:28




















          • $begingroup$
            I did draw an image similar to yours. I got similar values for inner and outer radii and could NOT get the text answer which is 124pi/5.
            $endgroup$
            – user163862
            Dec 29 '18 at 2:16










          • $begingroup$
            Can you assist me in understanding why there are 2 inner radii?
            $endgroup$
            – user163862
            Dec 29 '18 at 2:30










          • $begingroup$
            @user163862: The left side of the shaded area consists of a straight line from $(1,0)$ to $(1,1)$, and the part of the parabola between $(1,1)$ and $(4,2)$. The sharp bend in this inside edge corresponds to the switch between expressions.
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 3:28


















          $begingroup$
          I did draw an image similar to yours. I got similar values for inner and outer radii and could NOT get the text answer which is 124pi/5.
          $endgroup$
          – user163862
          Dec 29 '18 at 2:16




          $begingroup$
          I did draw an image similar to yours. I got similar values for inner and outer radii and could NOT get the text answer which is 124pi/5.
          $endgroup$
          – user163862
          Dec 29 '18 at 2:16












          $begingroup$
          Can you assist me in understanding why there are 2 inner radii?
          $endgroup$
          – user163862
          Dec 29 '18 at 2:30




          $begingroup$
          Can you assist me in understanding why there are 2 inner radii?
          $endgroup$
          – user163862
          Dec 29 '18 at 2:30












          $begingroup$
          @user163862: The left side of the shaded area consists of a straight line from $(1,0)$ to $(1,1)$, and the part of the parabola between $(1,1)$ and $(4,2)$. The sharp bend in this inside edge corresponds to the switch between expressions.
          $endgroup$
          – Henning Makholm
          Dec 29 '18 at 3:28






          $begingroup$
          @user163862: The left side of the shaded area consists of a straight line from $(1,0)$ to $(1,1)$, and the part of the parabola between $(1,1)$ and $(4,2)$. The sharp bend in this inside edge corresponds to the switch between expressions.
          $endgroup$
          – Henning Makholm
          Dec 29 '18 at 3:28













          0












          $begingroup$

          The inner radius would be $1$ for $0leq y<1$ and $y^2$ for $1leq yleq2$.



          The outer radius would be 4 for $0leq yleq2$



          Then, the set up for the volume would be
          $$V=piint_{0}^{1}(4^2-1^2)~dy+piint_{1}^{2}(4^2-y^4)~dy=frac{124pi}{5}$$
          enter image description here






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How on earth do you get those radiuses?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 0:43










          • $begingroup$
            I used the graph of $x=y^2$. Since the area is bounded between x=4 and x=1, I think the outer radius should be $4-y^2$, and the inner radius should be $1-y^2$. Did I do anything wrong?
            $endgroup$
            – Larry
            Dec 29 '18 at 0:49










          • $begingroup$
            I'm still quite baffled where you're measuring those radii. Are you imagining we're you're rotating about the line $x=4$? No, that doesn't make sense either. Perhaps you could draw a diagram to show more precisely where you get those distances from?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 0:58










          • $begingroup$
            I think I made a mistake, I'll edit my answer.
            $endgroup$
            – Larry
            Dec 29 '18 at 1:01






          • 1




            $begingroup$
            x @Larry: Um ... I can see what you're doing now, but it is quite wrong. "Radius" means distance from the axis of revolution, and the axis of revolution is the y-axis, not the red parabola! How would you rotate something about a curved axis anyway?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:23


















          0












          $begingroup$

          The inner radius would be $1$ for $0leq y<1$ and $y^2$ for $1leq yleq2$.



          The outer radius would be 4 for $0leq yleq2$



          Then, the set up for the volume would be
          $$V=piint_{0}^{1}(4^2-1^2)~dy+piint_{1}^{2}(4^2-y^4)~dy=frac{124pi}{5}$$
          enter image description here






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            How on earth do you get those radiuses?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 0:43










          • $begingroup$
            I used the graph of $x=y^2$. Since the area is bounded between x=4 and x=1, I think the outer radius should be $4-y^2$, and the inner radius should be $1-y^2$. Did I do anything wrong?
            $endgroup$
            – Larry
            Dec 29 '18 at 0:49










          • $begingroup$
            I'm still quite baffled where you're measuring those radii. Are you imagining we're you're rotating about the line $x=4$? No, that doesn't make sense either. Perhaps you could draw a diagram to show more precisely where you get those distances from?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 0:58










          • $begingroup$
            I think I made a mistake, I'll edit my answer.
            $endgroup$
            – Larry
            Dec 29 '18 at 1:01






          • 1




            $begingroup$
            x @Larry: Um ... I can see what you're doing now, but it is quite wrong. "Radius" means distance from the axis of revolution, and the axis of revolution is the y-axis, not the red parabola! How would you rotate something about a curved axis anyway?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:23
















          0












          0








          0





          $begingroup$

          The inner radius would be $1$ for $0leq y<1$ and $y^2$ for $1leq yleq2$.



          The outer radius would be 4 for $0leq yleq2$



          Then, the set up for the volume would be
          $$V=piint_{0}^{1}(4^2-1^2)~dy+piint_{1}^{2}(4^2-y^4)~dy=frac{124pi}{5}$$
          enter image description here






          share|cite|improve this answer











          $endgroup$



          The inner radius would be $1$ for $0leq y<1$ and $y^2$ for $1leq yleq2$.



          The outer radius would be 4 for $0leq yleq2$



          Then, the set up for the volume would be
          $$V=piint_{0}^{1}(4^2-1^2)~dy+piint_{1}^{2}(4^2-y^4)~dy=frac{124pi}{5}$$
          enter image description here







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 29 '18 at 2:11

























          answered Dec 29 '18 at 0:40









          LarryLarry

          2,41331129




          2,41331129












          • $begingroup$
            How on earth do you get those radiuses?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 0:43










          • $begingroup$
            I used the graph of $x=y^2$. Since the area is bounded between x=4 and x=1, I think the outer radius should be $4-y^2$, and the inner radius should be $1-y^2$. Did I do anything wrong?
            $endgroup$
            – Larry
            Dec 29 '18 at 0:49










          • $begingroup$
            I'm still quite baffled where you're measuring those radii. Are you imagining we're you're rotating about the line $x=4$? No, that doesn't make sense either. Perhaps you could draw a diagram to show more precisely where you get those distances from?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 0:58










          • $begingroup$
            I think I made a mistake, I'll edit my answer.
            $endgroup$
            – Larry
            Dec 29 '18 at 1:01






          • 1




            $begingroup$
            x @Larry: Um ... I can see what you're doing now, but it is quite wrong. "Radius" means distance from the axis of revolution, and the axis of revolution is the y-axis, not the red parabola! How would you rotate something about a curved axis anyway?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:23




















          • $begingroup$
            How on earth do you get those radiuses?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 0:43










          • $begingroup$
            I used the graph of $x=y^2$. Since the area is bounded between x=4 and x=1, I think the outer radius should be $4-y^2$, and the inner radius should be $1-y^2$. Did I do anything wrong?
            $endgroup$
            – Larry
            Dec 29 '18 at 0:49










          • $begingroup$
            I'm still quite baffled where you're measuring those radii. Are you imagining we're you're rotating about the line $x=4$? No, that doesn't make sense either. Perhaps you could draw a diagram to show more precisely where you get those distances from?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 0:58










          • $begingroup$
            I think I made a mistake, I'll edit my answer.
            $endgroup$
            – Larry
            Dec 29 '18 at 1:01






          • 1




            $begingroup$
            x @Larry: Um ... I can see what you're doing now, but it is quite wrong. "Radius" means distance from the axis of revolution, and the axis of revolution is the y-axis, not the red parabola! How would you rotate something about a curved axis anyway?
            $endgroup$
            – Henning Makholm
            Dec 29 '18 at 1:23


















          $begingroup$
          How on earth do you get those radiuses?
          $endgroup$
          – Henning Makholm
          Dec 29 '18 at 0:43




          $begingroup$
          How on earth do you get those radiuses?
          $endgroup$
          – Henning Makholm
          Dec 29 '18 at 0:43












          $begingroup$
          I used the graph of $x=y^2$. Since the area is bounded between x=4 and x=1, I think the outer radius should be $4-y^2$, and the inner radius should be $1-y^2$. Did I do anything wrong?
          $endgroup$
          – Larry
          Dec 29 '18 at 0:49




          $begingroup$
          I used the graph of $x=y^2$. Since the area is bounded between x=4 and x=1, I think the outer radius should be $4-y^2$, and the inner radius should be $1-y^2$. Did I do anything wrong?
          $endgroup$
          – Larry
          Dec 29 '18 at 0:49












          $begingroup$
          I'm still quite baffled where you're measuring those radii. Are you imagining we're you're rotating about the line $x=4$? No, that doesn't make sense either. Perhaps you could draw a diagram to show more precisely where you get those distances from?
          $endgroup$
          – Henning Makholm
          Dec 29 '18 at 0:58




          $begingroup$
          I'm still quite baffled where you're measuring those radii. Are you imagining we're you're rotating about the line $x=4$? No, that doesn't make sense either. Perhaps you could draw a diagram to show more precisely where you get those distances from?
          $endgroup$
          – Henning Makholm
          Dec 29 '18 at 0:58












          $begingroup$
          I think I made a mistake, I'll edit my answer.
          $endgroup$
          – Larry
          Dec 29 '18 at 1:01




          $begingroup$
          I think I made a mistake, I'll edit my answer.
          $endgroup$
          – Larry
          Dec 29 '18 at 1:01




          1




          1




          $begingroup$
          x @Larry: Um ... I can see what you're doing now, but it is quite wrong. "Radius" means distance from the axis of revolution, and the axis of revolution is the y-axis, not the red parabola! How would you rotate something about a curved axis anyway?
          $endgroup$
          – Henning Makholm
          Dec 29 '18 at 1:23






          $begingroup$
          x @Larry: Um ... I can see what you're doing now, but it is quite wrong. "Radius" means distance from the axis of revolution, and the axis of revolution is the y-axis, not the red parabola! How would you rotate something about a curved axis anyway?
          $endgroup$
          – Henning Makholm
          Dec 29 '18 at 1:23




















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