How can I write the radius equation for the disk method if the axis of revolution intersects the area between...
$begingroup$
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
asked Sep 13 '15 at 20:05
Joe StavitskyJoe Stavitsky
2461317
$endgroup$
add a comment |
$begingroup$
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
asked Sep 13 '15 at 20:05
Joe StavitskyJoe Stavitsky
2461317
$endgroup$
$begingroup$
It's not clear what area you are rotating....
$endgroup$
– David Quinn
Sep 13 '15 at 20:13
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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
add a comment |
$begingroup$
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
asked Sep 13 '15 at 20:05
Joe StavitskyJoe Stavitsky
2461317
$endgroup$
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
asked Sep 13 '15 at 20:05
Joe StavitskyJoe Stavitsky
2461317
asked Sep 13 '15 at 20:05
Joe StavitskyJoe Stavitsky
2461317
edited Sep 13 '15 at 20:22
Joe Stavitsky
asked Sep 13 '15 at 20:05
Joe StavitskyJoe Stavitsky
2461317
asked Sep 13 '15 at 20:05
Joe StavitskyJoe Stavitsky
2461317
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $y=-2$ ...
answered Sep 13 '15 at 20:28
zahbazzahbaz
8,30421937
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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1 Answer
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oldest
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$begingroup$
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 $y=-2$ ...
answered Sep 13 '15 at 20:28
zahbazzahbaz
8,30421937
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $y=-2$ ...
answered Sep 13 '15 at 20:28
zahbazzahbaz
8,30421937
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $y=-2$ ...
answered Sep 13 '15 at 20:28
zahbazzahbaz
8,30421937
$endgroup$
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 $y=-2$ ...
answered Sep 13 '15 at 20:28
zahbazzahbaz
8,30421937
edited Sep 13 '15 at 20:40
answered Sep 13 '15 at 20:28
zahbazzahbaz
8,30421937
answered Sep 13 '15 at 20:28
zahbazzahbaz
8,30421937
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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