Revolve the region under the curve y =√x on the interval [0, 4] about the x-axis and
find the volume of the resulting solid of revolution.
YET, NOW, I want to do in respect to y
The Attempt at a Solution
I did dx and I got 8pi which is corrected.
So I decided to do it with dy.
I looked at the graph and I saw the height would be the same (well because of the problem's nature). so a and b will stay the same
integral of pi*r^2
we know r is y^2 (since y^2 = x)
integral of pi * (y^2)^2 from 1 to 4
but I did not get 8pi