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 January 29th, 2013, 07:50 PM #1 Member   Joined: Oct 2012 Posts: 94 Thanks: 0 area and volume between curves 4. Find the volume of the solid generated by rotating the region in the fi rst quadrant bounded by y = x^2, the y-axis, and y = 9 around x = 0. So, since we are rotating around the y-axis, we need to put everything in terms of y, right? and we need to split the problem into two different integrals, right? one for 0 - x^2 and one for 9 - x^2. I don't understand how to figure out the upper bound for 0 - x^2 and the lower bound for 9 - x^2. Is it 3? I know that the lower bound of the first integral is zero and the upper bound for the second is 9. A second question. b) The solid obtained by rotating the region bounded by y =(4-x)^1/2, the x-axis and the y-axis around the line x = 0. So, again, we are rotating around the y axis, so we convert everything to terms of y. Why is the upper bound 4? Shouldn't that be the upper bound if we were rotating around the y=0 and are working in terms of x? In terms of y, shouldn't the upper bound be 2? That is how it is when when finding areas between curves, so I don't really understand why there's a difference.
 January 29th, 2013, 08:50 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: area and volume between curves The first problem: The disk method: First, compute the volume of an arbitrary disk: $dV=\pi r^2\,dy$ $r=x\,\therefore\,r^2=x^2=y$ and we have: $dV=\pi y\,dy$ Summing the disks by integration, we find: $V=\pi\int_0\,^9 y\,dy=\frac{\pi}{2}$y^2$_0^9=\frac{81\pi}{2}$ The shell method: The volume of an arbitrary shell is: $dV=2\pi rh\,dx$ $r=x$ $h=9-x^2$ and we have: $dV=2\pi x(9-x^2)\,dx=2\pi$$9x-x^3$$\,dx$ Summing the disks by integration, we find: $V=2\pi\int_0\,^3 9x-x^3\,dx=\frac{\pi}{2}$18x^2-x^4$_0^3=\frac{81\pi}{2}$ Now, I will let you try the second one, and post your work so we can see if you have the right technique.
 January 29th, 2013, 09:30 PM #3 Member   Joined: Oct 2012 Posts: 94 Thanks: 0 Re: area and volume between curves I am still confused. My second question had more to do with why we were evaluating from 0 to 4 rather than 0 to 2? My professor says we evaluate pi * (4-y^2)^2 from zero to 4. I understand the integration. But, I thought, in order to find the upper bound we were to set 4-y^2 equal to zero (in this case), which equals 2. If I look at a graph of these function, 2 is the max. I could understand if we were evaluating int terms of x, because the domain of the area we are analyzing is from 0 to 4. But, since we are looking at it in terms of y, shouldn't it be 2? Thank you.
 January 29th, 2013, 09:39 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: area and volume between curves If you are going to use the disk method, then the limits are from 0 to 2, but if you are going to use the shell method, then the limits are from 0 to 4.
 January 29th, 2013, 09:41 PM #5 Member   Joined: Oct 2012 Posts: 94 Thanks: 0 Re: area and volume between curves Could you explain that a little? It just doesn't seem consistent with all of the other problems I've done, so I know I must be missing something.
January 29th, 2013, 10:06 PM   #6
Senior Member

Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,211
Thanks: 521

Math Focus: Calculus/ODEs
Re: area and volume between curves

Here is a plot of the region to be revolved:

[attachment=0:104huzrh]solidrevolution.jpg[/attachment:104huzrh]

The horizontal red line represents the radius of an arbitrary disk. Notice the horizontal lines will run from y = 0 to y = 2.

The radius of the disk is $x=4-y^2$, and the thickness is $dy$.

The vertical green line represents the height of an arbitrary shell. Notice the vertical lines will run from x = 0 to x = 4.

The radius of the cylindrical shell is $x$, the height is $y=\sqrt{4-x}$ and the thickness is $dx$.
Attached Images
 solidrevolution.jpg (10.4 KB, 71 views)

 January 29th, 2013, 10:23 PM #7 Member   Joined: Oct 2012 Posts: 94 Thanks: 0 Re: area and volume between curves That makes perfect sense! It's what I've been doing. My teacher must have made a typo, because as you can see, she placed the disk radius with the shell limits! oh my gosh. i have been staring at this for an hour trying to figure out what I've been doing wrong. 0_o Thanks!

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