January 29th, 2013, 08:50 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 first quadrant bounded by y = x^2, the yaxis, and y = 9 around x = 0. So, since we are rotating around the yaxis, 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 =(4x)^1/2, the xaxis and the yaxis 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, 09:50 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 Math Focus: Calculus/ODEs  Re: area and volume between curves
The first problem: The disk method: First, compute the volume of an arbitrary disk: and we have: Summing the disks by integration, we find: The shell method: The volume of an arbitrary shell is: and we have: Summing the disks by integration, we find: 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, 10:30 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 * (4y^2)^2 from zero to 4. I understand the integration. But, I thought, in order to find the upper bound we were to set 4y^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, 10:39 PM  #4 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 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, 10:41 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, 11:06 PM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 520 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 , and the thickness is . 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 , the height is and the thickness is . 
January 29th, 2013, 11:23 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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