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March 29th, 2015, 05:26 PM  #1 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36  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 xaxis is $\displaystyle 2\pi (1+2\ln 3)$ units^3.
Last edited by skipjack; March 29th, 2015 at 08:02 PM. 
March 29th, 2015, 05:56 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs 
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: $\displaystyle V=4\pi\int_1^3 x\frac{1}{x}\,dx=4\pi\left[\frac{1}{2}x^2\ln(x)\right]_1^3=4\pi\left(4\ln(3)\right)$ 
March 29th, 2015, 06:03 PM  #3 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,807 Thanks: 1045 Math Focus: Elementary mathematics and beyond 
Why would it be a "full revolution"?

March 29th, 2015, 06:17 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs  
March 29th, 2015, 06:21 PM  #5 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 
but the question answer is not same with yours? I'm confusing

March 29th, 2015, 06:43 PM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs  Well, let's see what we get using the shell method. The volume of an element is: $\displaystyle dV=2\pi y(3x)\,dy$ 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: $\displaystyle V=2\pi\left(\int_{\frac{2}{\sqrt{3}}}^2 3y\frac{4}{y}\,dy+\int_2^{2\sqrt{3}}3y\frac{1}{4}y^3\,dy\right)$ $\displaystyle V=2\pi\left(82\ln(3)\right)=4\pi\left(4\ln(3)\right)$ 
March 29th, 2015, 06:53 PM  #7 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 
So that means the question is wrong. No wonder I couldn't find the answer .

March 29th, 2015, 08:10 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 18,965 Thanks: 1606 
The question is right (albeit very poorly worded) and MarkFL is wrong (due to using an unintended interpretation of the question).

March 29th, 2015, 08:33 PM  #9 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36  
March 29th, 2015, 08:37 PM  #10 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs 
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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