Find the volume of the solid formed by revolving the region bounded by the graphs of y = x^2, x = 4, and y = 1 about the y-axis.

Question

I'm supposed to use this equation right? piR^2-x^r^2. and then change y=x^2 to x=sqrty. That's all I got.

Answer 1

as you say, you have to think of the solid as a stack of washers, of thickness dy. So,

v = ∫[1,16] π(R^2-r^2) dy
where R = 4 and r = x = √y. So,
v = ∫[1,16] π(16-y) dy
= 225π/2

As a check on your answer, think of the solid as a set of nested cylinders of thickness dx. Then you have

v = ∫[1,4] 2πrh dx
where r = x and h = y-1 = x^2-1. Thus,
v = ∫[1,4] 2πx(x^2-1) dx
= 225π/2

Answer 2

225 pi/2 is right. i just did this question.