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 x-axis.

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^2-r^2)h

where the height is dx and the radii are the distance from the x-axis 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 x-axis, 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^2-r^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 x-axis. 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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