# Volume of Revolution By Shell Method.

1. Feb 24, 2016

### OmniNewton

1. The problem statement, all variables and given/known data
Find the volume of the solid obtained by rotating y=x^2 and x = y^2 about y=2

2. Relevant equations
V= 2πrh

3. The attempt at a solution
When I had constructed the graph I determined the following:
r= 2-(y(1/2))
h= 1-y2

after converting V into an integral I applied fundamental theorem of calculus.
2-2y^2-y^(1/2) + y^(5/2)
evaluating the expression from lower limit =0 to upper limit = 1

I obtained a value with a denominator of 21.

The final answer I should receive is 31π/30 (I proved this by disk method and this is the stated answer).

2. Feb 24, 2016

### Staff: Mentor

I have r = 2 - y and h = $y^{1/2} - y^2$.
I get 31π/30 for the volume.

3. Feb 24, 2016

### OmniNewton

Oh ok I see how you got that. Let me give that a try.

4. Feb 24, 2016

### OmniNewton

Thank you so much it worked out! I was at that problem for a good two hours. I erased so much my page nearly ripped. After taking a break to work on another problem and then coming back with your guidance I was able to figure it out.

5. Feb 24, 2016

### Staff: Mentor

I hope you're drawing sketches of 1) the region being revolved, with the two curves and important points labeled, and 2) the solid of revolution. Based on what you showed, I think you did, but sometimes people think they can save time by skipping that step. It's almost always a false economy...

6. Feb 24, 2016

### OmniNewton

Yes, my steps for these problems are.

1. draw the region
2. find necessary point of intersections
3. split the axis of rotation for disk method or split the perpendicular axis of rotation for shell method
4. draw atleast one disk or shell (more may be needed)
5. then work from there.

My problem is sometimes I still have a hard time noticing key things like radius and height (getting better with practice)

7. Feb 24, 2016

### Staff: Mentor

Very good. I think the problem you had was not realizing what the appropriate x- or y-coordinates were for writing the radius and the width of the area element. You'll get better at this with practice. Other than that, you seem to be on the right track.

8. Feb 24, 2016

### OmniNewton

Yes, that makes sense. I believe that is what happened