Posted by Jackie on Wednesday, December 4, 2013 at 2:53pm.
Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.
y = 10 x and y = 5 x^2 about y =0
Find the volume of the solid obtained by rotating the region bounded by y=8 x^2, x = 1, and y = 0 , about the xaxis.
I saw some other people were having trouble too and looked at those, but it makes no sense to me. Help? :(

Calculus I don't understand  Steve, Wednesday, December 4, 2013 at 3:41pm
ok. for volume generated by rotating around y=0, you can use discs or shells. The volume of a disc, or washer with a hole in it, is
v=π(R^2r^2)h
where the height is dx and the radii are the distance from the xaxis to the two curves.
The upper curve is y=10x and the lower curve is y=5x^2. If you take a thin slice of the area and rotate it around the xaxis, you will get a washer.
Since the two curves intersect at (0,0) and (2,20), we integrate along x from 0 to 2.
v = ∫[0,2] π(R^2r^2) dx
= π∫[0,2] (10x)^2  (5x^2)^2 dx
= π∫[0,2] 100x^2  25x^4 dx
= π(100/3 x^3  5x^5) [0,2]
= π(800/3  160)
= 320/3 π
Now, you can also use shells. Recall that the volume of a thin cylinder of thickness dy is just
v = 2πrh dy
Now, in this case, the radius is just y, since we are rotating around the xaxis. The height h is the distance between the two curves
x = √(y/5) and
x = y/10
So, our shell volume is
v = ∫[0,20] 2πy(√(y/5)y/10)
= 2π∫[0,20] 1/√5 y^(3/2)  y^2/10 dy
= 2π(2/5√5 y^(5/2)  y^3/30) [0,20]
= 2π(2/5√5 * 400√20  800/3)
= 2π(320  800/3)
= 320/3 π
Now see whether you can set up the other one. As you can probably see, using discs is easier in this case.

Calculus I don't understand  Anonymous, Wednesday, December 4, 2013 at 7:10pm
Thank you so much that makes sense more sense to me! :)
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