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March 29th, 2015, 05:26 PM   #1
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Application of integration. (Volume)

Find the points of intersection of the curves $\displaystyle y^2=4x$ and $\displaystyle y^2=\frac{4}{x}$. Show that the volume of the solid generated when the region bounded by the curves and the straight line $\displaystyle x=3$ is revolved through $\displaystyle \pi$ radians about the x-axis is $\displaystyle 2\pi (1+2\ln 3)$ units^3.

Last edited by skipjack; March 29th, 2015 at 08:02 PM.
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March 29th, 2015, 05:56 PM   #2
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This will be equivalent to revolving the region bounded by:

$\displaystyle y=2\sqrt{x}$

$\displaystyle y=\frac{2}{\sqrt{x}}$

$\displaystyle x=3$

about the $x$-axis a full revolution. Using the washer method, we find an element of the volume is:

$\displaystyle dV=4\pi\left(x-\frac{1}{x}\right)\,dx$

Hence, the volume is:

$\displaystyle V=4\pi\int_1^3 x-\frac{1}{x}\,dx=4\pi\left[\frac{1}{2}x^2-\ln(x)\right]_1^3=4\pi\left(4-\ln(3)\right)$
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March 29th, 2015, 06:03 PM   #3
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Why would it be a "full revolution"?
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March 29th, 2015, 06:17 PM   #4
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Originally Posted by greg1313 View Post
Why would it be a "full revolution"?
The region in question is symmetric across the $x$-axis, and so revolving the top region a full revolution is equivalent to revolving both regions a half-revolution.
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March 29th, 2015, 06:21 PM   #5
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but the question answer is not same with yours? I'm confusing
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March 29th, 2015, 06:43 PM   #6
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but the question answer is not same with yours? I'm confusing
Well, let's see what we get using the shell method. The volume of an element is:

$\displaystyle dV=2\pi y(3-x)\,dy$

Now when $\displaystyle \frac{2}{\sqrt{3}}\le y\le2$ we have:

$\displaystyle x=\frac{4}{y^2}$

And when $\displaystyle 2\le y\le2\sqrt{3}$ we have:

$\displaystyle x=\frac{y^2}{4}$

And so we find:

$\displaystyle V=2\pi\left(\int_{\frac{2}{\sqrt{3}}}^2 3y-\frac{4}{y}\,dy+\int_2^{2\sqrt{3}}3y-\frac{1}{4}y^3\,dy\right)$

$\displaystyle V=2\pi\left(8-2\ln(3)\right)=4\pi\left(4-\ln(3)\right)$
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March 29th, 2015, 06:53 PM   #7
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So that means the question is wrong. No wonder I couldn't find the answer .
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March 29th, 2015, 08:10 PM   #8
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The question is right (albeit very poorly worded) and MarkFL is wrong (due to using an unintended interpretation of the question).
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March 29th, 2015, 08:33 PM   #9
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Quote:
Originally Posted by skipjack View Post
The question is right (albeit very poorly worded) and MarkFL is wrong (due to using an unintended interpretation of the question).
Can you show me the solution?
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March 29th, 2015, 08:37 PM   #10
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Okay, using the other region meeting the given conditions, we find:

$\displaystyle V=4\pi\left(\int_0^1 x\,dx+\int_1^3\frac{1}{x} \right)=2\pi\left(1+2\ln(3)\right)$
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