Let R be the region between the graph of y=x3 and the x-axis, and between x=0 and x=1.

Question

Compute the volume of the region obtained by revolving R around the y-axis

Answer 1

with discs (washers),
v = ∫[0,1] π (R^2-r^2) dy
where r=x and R=1
v = π∫[0,1] (1 - y^(2/3)) dy
= π (y - 3/5 y^(5/3)) [0,1]
= 2/5 π

with shells,
v = ∫[0,1] 2π rh dx
where r=x and h=y
v = 2π∫[0,1] x*x^3 dx
= 2π (1/5 x^5) [0,1]
= 2/5 π

Answer 2

First off, thank you answering my question.

I just have one question. So does it matter which method we use?

I am just trying to find a better way of understanding this problem from a visual perspective.

Answer 3

generally it doesn't matter. It depends on whether it is easier to express y as a function of x, or vice-versa. For example if y=e^x + 4(x^2-1) then it's pretty tough to come up with y^-1

In this case, it was easy.