My Math Forum Application of integration. (Volume)

 Calculus Calculus Math Forum

 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 x-axis 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
Quote:
 Originally Posted by greg1313 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.

 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
Quote:
 Originally Posted by jiasyuen 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:

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

$\displaystyle V=2\pi\left(8-2\ln(3)\right)=4\pi\left(4-\ln(3)\right)$

 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

Quote:
 Originally Posted by skipjack 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?

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

 Tags application, integration, volume

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post David_Lete Calculus 3 January 31st, 2010 06:47 PM David_Lete Calculus 1 January 22nd, 2010 04:01 PM NiCeBoY Calculus 2 May 17th, 2009 01:03 PM arun Calculus 1 September 27th, 2008 02:59 PM arun Calculus 0 September 27th, 2008 08:16 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top