How can I write the radius equation for the disk method if the axis of revolution intersects the area between...












0












$begingroup$


A specific example would be revolving the area between $x^2-5$ and $5x$ below the $x$-axis about $y=-2$



PS - in general, I am assuming that revolving about any other horizontal or vertical line entails subtracting the relevant variable from the constant value - is that accurate?






calculus definite-integrals applications







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$endgroup$












  • $begingroup$
    It's not clear what area you are rotating....
    $endgroup$
    – David Quinn
    Sep 13 '15 at 20:13










  • $begingroup$
    There is no region between $y= x^2$ and $y=5x$ that lies below the $x$-axis? You're not rotating any region?
    $endgroup$
    – DJS
    Sep 13 '15 at 20:15












  • $begingroup$
    He's rotating the region between $x^2$ and $5x$ from $xin [0,5]$ about $y=-2$.
    $endgroup$
    – zahbaz
    Sep 13 '15 at 20:16












  • $begingroup$
    @zahbaz - He stated "the area between $x^2$ and $5x$ below the $x$-axis "? There is no such region. I agree that I think he meant what you said, but his wording was incorrect.
    $endgroup$
    – DJS
    Sep 13 '15 at 20:17












  • $begingroup$
    sorry - typo fixed
    $endgroup$
    – Joe Stavitsky
    Sep 13 '15 at 20:22
















0












$begingroup$


A specific example would be revolving the area between $x^2-5$ and $5x$ below the $x$-axis about $y=-2$



PS - in general, I am assuming that revolving about any other horizontal or vertical line entails subtracting the relevant variable from the constant value - is that accurate?






calculus definite-integrals applications







share|cite|improve this question











$endgroup$












  • $begingroup$
    It's not clear what area you are rotating....
    $endgroup$
    – David Quinn
    Sep 13 '15 at 20:13










  • $begingroup$
    There is no region between $y= x^2$ and $y=5x$ that lies below the $x$-axis? You're not rotating any region?
    $endgroup$
    – DJS
    Sep 13 '15 at 20:15












  • $begingroup$
    He's rotating the region between $x^2$ and $5x$ from $xin [0,5]$ about $y=-2$.
    $endgroup$
    – zahbaz
    Sep 13 '15 at 20:16












  • $begingroup$
    @zahbaz - He stated "the area between $x^2$ and $5x$ below the $x$-axis "? There is no such region. I agree that I think he meant what you said, but his wording was incorrect.
    $endgroup$
    – DJS
    Sep 13 '15 at 20:17












  • $begingroup$
    sorry - typo fixed
    $endgroup$
    – Joe Stavitsky
    Sep 13 '15 at 20:22














0












0








0





$begingroup$


A specific example would be revolving the area between $x^2-5$ and $5x$ below the $x$-axis about $y=-2$



PS - in general, I am assuming that revolving about any other horizontal or vertical line entails subtracting the relevant variable from the constant value - is that accurate?






calculus definite-integrals applications







share|cite|improve this question











$endgroup$




A specific example would be revolving the area between $x^2-5$ and $5x$ below the $x$-axis about $y=-2$



PS - in general, I am assuming that revolving about any other horizontal or vertical line entails subtracting the relevant variable from the constant value - is that accurate?






calculus definite-integrals applications




calculus definite-integrals applications






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 13 '15 at 20:22







Joe Stavitsky

















asked Sep 13 '15 at 20:05









Joe StavitskyJoe Stavitsky

2461317




2461317












  • $begingroup$
    It's not clear what area you are rotating....
    $endgroup$
    – David Quinn
    Sep 13 '15 at 20:13










  • $begingroup$
    There is no region between $y= x^2$ and $y=5x$ that lies below the $x$-axis? You're not rotating any region?
    $endgroup$
    – DJS
    Sep 13 '15 at 20:15












  • $begingroup$
    He's rotating the region between $x^2$ and $5x$ from $xin [0,5]$ about $y=-2$.
    $endgroup$
    – zahbaz
    Sep 13 '15 at 20:16












  • $begingroup$
    @zahbaz - He stated "the area between $x^2$ and $5x$ below the $x$-axis "? There is no such region. I agree that I think he meant what you said, but his wording was incorrect.
    $endgroup$
    – DJS
    Sep 13 '15 at 20:17












  • $begingroup$
    sorry - typo fixed
    $endgroup$
    – Joe Stavitsky
    Sep 13 '15 at 20:22


















  • $begingroup$
    It's not clear what area you are rotating....
    $endgroup$
    – David Quinn
    Sep 13 '15 at 20:13










  • $begingroup$
    There is no region between $y= x^2$ and $y=5x$ that lies below the $x$-axis? You're not rotating any region?
    $endgroup$
    – DJS
    Sep 13 '15 at 20:15












  • $begingroup$
    He's rotating the region between $x^2$ and $5x$ from $xin [0,5]$ about $y=-2$.
    $endgroup$
    – zahbaz
    Sep 13 '15 at 20:16












  • $begingroup$
    @zahbaz - He stated "the area between $x^2$ and $5x$ below the $x$-axis "? There is no such region. I agree that I think he meant what you said, but his wording was incorrect.
    $endgroup$
    – DJS
    Sep 13 '15 at 20:17












  • $begingroup$
    sorry - typo fixed
    $endgroup$
    – Joe Stavitsky
    Sep 13 '15 at 20:22
















$begingroup$
It's not clear what area you are rotating....
$endgroup$
– David Quinn
Sep 13 '15 at 20:13




$begingroup$
It's not clear what area you are rotating....
$endgroup$
– David Quinn
Sep 13 '15 at 20:13












$begingroup$
There is no region between $y= x^2$ and $y=5x$ that lies below the $x$-axis? You're not rotating any region?
$endgroup$
– DJS
Sep 13 '15 at 20:15






$begingroup$
There is no region between $y= x^2$ and $y=5x$ that lies below the $x$-axis? You're not rotating any region?
$endgroup$
– DJS
Sep 13 '15 at 20:15














$begingroup$
He's rotating the region between $x^2$ and $5x$ from $xin [0,5]$ about $y=-2$.
$endgroup$
– zahbaz
Sep 13 '15 at 20:16






$begingroup$
He's rotating the region between $x^2$ and $5x$ from $xin [0,5]$ about $y=-2$.
$endgroup$
– zahbaz
Sep 13 '15 at 20:16














$begingroup$
@zahbaz - He stated "the area between $x^2$ and $5x$ below the $x$-axis "? There is no such region. I agree that I think he meant what you said, but his wording was incorrect.
$endgroup$
– DJS
Sep 13 '15 at 20:17






$begingroup$
@zahbaz - He stated "the area between $x^2$ and $5x$ below the $x$-axis "? There is no such region. I agree that I think he meant what you said, but his wording was incorrect.
$endgroup$
– DJS
Sep 13 '15 at 20:17














$begingroup$
sorry - typo fixed
$endgroup$
– Joe Stavitsky
Sep 13 '15 at 20:22




$begingroup$
sorry - typo fixed
$endgroup$
– Joe Stavitsky
Sep 13 '15 at 20:22










1 Answer
1






active

oldest

votes


















0












$begingroup$

Start with drawing a picture. In this case, think washer or donut instead of disk. But first, think of how you'd calculate the area of a donut. You'd say



$$A = pi R^2 - pi r^2$$



where $R$ is the outer and $r$ is the inner radius. This problem is similar, except we're stacking all these washers as we march from $x=0$ to $x=5$ where the two curves intersect.



The outer radius goes from $y=-2$ to the top curve, $5x$ and the inner radius goes to $x^2$. Draw this! You ought to see



$$R = 2 + 5x$$
$$r = 2 + x^2$$



Next, set up your integral



$$pi int_{x=0}^5 R^2 - r^2 dx$$




It looks like the problem changed to the curves $x^2 -5$ and $5x$. This is now unclear. Do you want to rotate the area bounded by those two curves and the x axis, $y=0$? If so, the rotated region will overlap itself when rotated about $y=-2$ ...






share|cite|improve this answer











$endgroup$













  • $begingroup$
    the intersection was intentional - the question was how to write R and r when the xis of rotation intersects one of the functions
    $endgroup$
    – Joe Stavitsky
    Sep 13 '15 at 22:36










  • $begingroup$
    Volumes of rotation are a physical idea, and physically speaking, if we the rotate the entire bounded region completely about $y=-2$, we would rotate the region back onto itself after rotating through $pi$. That's not a well-defined volume. So it depends on how you define the problem. Do we stop rotating after $pi$? If not, do we select the exterior surface to define our volume? Or, do we create two volumes - one for the rotated region below $y=-2$ and one for that above? It's up to interpretation. desmos.com/calculator/uoshfogy7q
    $endgroup$
    – zahbaz
    Sep 14 '15 at 0:00













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1 Answer
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1 Answer
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active

oldest

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active

oldest

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0












$begingroup$

Start with drawing a picture. In this case, think washer or donut instead of disk. But first, think of how you'd calculate the area of a donut. You'd say



$$A = pi R^2 - pi r^2$$



where $R$ is the outer and $r$ is the inner radius. This problem is similar, except we're stacking all these washers as we march from $x=0$ to $x=5$ where the two curves intersect.



The outer radius goes from $y=-2$ to the top curve, $5x$ and the inner radius goes to $x^2$. Draw this! You ought to see



$$R = 2 + 5x$$
$$r = 2 + x^2$$



Next, set up your integral



$$pi int_{x=0}^5 R^2 - r^2 dx$$




It looks like the problem changed to the curves $x^2 -5$ and $5x$. This is now unclear. Do you want to rotate the area bounded by those two curves and the x axis, $y=0$? If so, the rotated region will overlap itself when rotated about $y=-2$ ...






share|cite|improve this answer











$endgroup$













  • $begingroup$
    the intersection was intentional - the question was how to write R and r when the xis of rotation intersects one of the functions
    $endgroup$
    – Joe Stavitsky
    Sep 13 '15 at 22:36










  • $begingroup$
    Volumes of rotation are a physical idea, and physically speaking, if we the rotate the entire bounded region completely about $y=-2$, we would rotate the region back onto itself after rotating through $pi$. That's not a well-defined volume. So it depends on how you define the problem. Do we stop rotating after $pi$? If not, do we select the exterior surface to define our volume? Or, do we create two volumes - one for the rotated region below $y=-2$ and one for that above? It's up to interpretation. desmos.com/calculator/uoshfogy7q
    $endgroup$
    – zahbaz
    Sep 14 '15 at 0:00


















0












$begingroup$

Start with drawing a picture. In this case, think washer or donut instead of disk. But first, think of how you'd calculate the area of a donut. You'd say



$$A = pi R^2 - pi r^2$$



where $R$ is the outer and $r$ is the inner radius. This problem is similar, except we're stacking all these washers as we march from $x=0$ to $x=5$ where the two curves intersect.



The outer radius goes from $y=-2$ to the top curve, $5x$ and the inner radius goes to $x^2$. Draw this! You ought to see



$$R = 2 + 5x$$
$$r = 2 + x^2$$



Next, set up your integral



$$pi int_{x=0}^5 R^2 - r^2 dx$$




It looks like the problem changed to the curves $x^2 -5$ and $5x$. This is now unclear. Do you want to rotate the area bounded by those two curves and the x axis, $y=0$? If so, the rotated region will overlap itself when rotated about $y=-2$ ...






share|cite|improve this answer











$endgroup$













  • $begingroup$
    the intersection was intentional - the question was how to write R and r when the xis of rotation intersects one of the functions
    $endgroup$
    – Joe Stavitsky
    Sep 13 '15 at 22:36










  • $begingroup$
    Volumes of rotation are a physical idea, and physically speaking, if we the rotate the entire bounded region completely about $y=-2$, we would rotate the region back onto itself after rotating through $pi$. That's not a well-defined volume. So it depends on how you define the problem. Do we stop rotating after $pi$? If not, do we select the exterior surface to define our volume? Or, do we create two volumes - one for the rotated region below $y=-2$ and one for that above? It's up to interpretation. desmos.com/calculator/uoshfogy7q
    $endgroup$
    – zahbaz
    Sep 14 '15 at 0:00
















0












0








0





$begingroup$

Start with drawing a picture. In this case, think washer or donut instead of disk. But first, think of how you'd calculate the area of a donut. You'd say



$$A = pi R^2 - pi r^2$$



where $R$ is the outer and $r$ is the inner radius. This problem is similar, except we're stacking all these washers as we march from $x=0$ to $x=5$ where the two curves intersect.



The outer radius goes from $y=-2$ to the top curve, $5x$ and the inner radius goes to $x^2$. Draw this! You ought to see



$$R = 2 + 5x$$
$$r = 2 + x^2$$



Next, set up your integral



$$pi int_{x=0}^5 R^2 - r^2 dx$$




It looks like the problem changed to the curves $x^2 -5$ and $5x$. This is now unclear. Do you want to rotate the area bounded by those two curves and the x axis, $y=0$? If so, the rotated region will overlap itself when rotated about $y=-2$ ...






share|cite|improve this answer











$endgroup$



Start with drawing a picture. In this case, think washer or donut instead of disk. But first, think of how you'd calculate the area of a donut. You'd say



$$A = pi R^2 - pi r^2$$



where $R$ is the outer and $r$ is the inner radius. This problem is similar, except we're stacking all these washers as we march from $x=0$ to $x=5$ where the two curves intersect.



The outer radius goes from $y=-2$ to the top curve, $5x$ and the inner radius goes to $x^2$. Draw this! You ought to see



$$R = 2 + 5x$$
$$r = 2 + x^2$$



Next, set up your integral



$$pi int_{x=0}^5 R^2 - r^2 dx$$




It looks like the problem changed to the curves $x^2 -5$ and $5x$. This is now unclear. Do you want to rotate the area bounded by those two curves and the x axis, $y=0$? If so, the rotated region will overlap itself when rotated about $y=-2$ ...







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 13 '15 at 20:40

























answered Sep 13 '15 at 20:28









zahbazzahbaz

8,30421937




8,30421937












  • $begingroup$
    the intersection was intentional - the question was how to write R and r when the xis of rotation intersects one of the functions
    $endgroup$
    – Joe Stavitsky
    Sep 13 '15 at 22:36










  • $begingroup$
    Volumes of rotation are a physical idea, and physically speaking, if we the rotate the entire bounded region completely about $y=-2$, we would rotate the region back onto itself after rotating through $pi$. That's not a well-defined volume. So it depends on how you define the problem. Do we stop rotating after $pi$? If not, do we select the exterior surface to define our volume? Or, do we create two volumes - one for the rotated region below $y=-2$ and one for that above? It's up to interpretation. desmos.com/calculator/uoshfogy7q
    $endgroup$
    – zahbaz
    Sep 14 '15 at 0:00




















  • $begingroup$
    the intersection was intentional - the question was how to write R and r when the xis of rotation intersects one of the functions
    $endgroup$
    – Joe Stavitsky
    Sep 13 '15 at 22:36










  • $begingroup$
    Volumes of rotation are a physical idea, and physically speaking, if we the rotate the entire bounded region completely about $y=-2$, we would rotate the region back onto itself after rotating through $pi$. That's not a well-defined volume. So it depends on how you define the problem. Do we stop rotating after $pi$? If not, do we select the exterior surface to define our volume? Or, do we create two volumes - one for the rotated region below $y=-2$ and one for that above? It's up to interpretation. desmos.com/calculator/uoshfogy7q
    $endgroup$
    – zahbaz
    Sep 14 '15 at 0:00


















$begingroup$
the intersection was intentional - the question was how to write R and r when the xis of rotation intersects one of the functions
$endgroup$
– Joe Stavitsky
Sep 13 '15 at 22:36




$begingroup$
the intersection was intentional - the question was how to write R and r when the xis of rotation intersects one of the functions
$endgroup$
– Joe Stavitsky
Sep 13 '15 at 22:36












$begingroup$
Volumes of rotation are a physical idea, and physically speaking, if we the rotate the entire bounded region completely about $y=-2$, we would rotate the region back onto itself after rotating through $pi$. That's not a well-defined volume. So it depends on how you define the problem. Do we stop rotating after $pi$? If not, do we select the exterior surface to define our volume? Or, do we create two volumes - one for the rotated region below $y=-2$ and one for that above? It's up to interpretation. desmos.com/calculator/uoshfogy7q
$endgroup$
– zahbaz
Sep 14 '15 at 0:00






$begingroup$
Volumes of rotation are a physical idea, and physically speaking, if we the rotate the entire bounded region completely about $y=-2$, we would rotate the region back onto itself after rotating through $pi$. That's not a well-defined volume. So it depends on how you define the problem. Do we stop rotating after $pi$? If not, do we select the exterior surface to define our volume? Or, do we create two volumes - one for the rotated region below $y=-2$ and one for that above? It's up to interpretation. desmos.com/calculator/uoshfogy7q
$endgroup$
– zahbaz
Sep 14 '15 at 0:00




















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