Calculus I don't understand
posted by Jackie on .
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? :(
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
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.
Thank you so much that makes sense more sense to me! :)