# Solid of revolution.

## Main Question or Discussion Point

Hi,
I have a question concerning solid of revolution.
The bowl-shaped volume formed by rotating the area circumscribed between y=bcosh(1) and y=bcosh(x/a) around the y axis was given to us by the instructor as pi*b*int [x^2*d(cosh(x/a))] between 0 and a.
My question is why are the integration boundaries not -a and a, but 0 and a, OR, alternatively, why wasn't the final answer then multiplied by 2?
When the same area is rotated around the x axis the integration boundaries are indeed -a and a. Why aren't the boundaries similar in both cases?

tiny-tim
Homework Helper
hi peripatein! The bowl-shaped volume formed by rotating the area circumscribed between y=bcosh(1) and y=bcosh(x/a) around the y axis was given to us by the instructor as pi*b*int [x^2*d(cosh(x/a))] between 0 and a.
this is very confusing do you mean the curve y = b cosh(x/a) from x = 0 (y = b) to x = a (y = b cosh(1)) ?

then your πx2dy formula gives you the whole horizontal disc of height dy, so there's no need to multiply it by 2 Hi tiny-tim,
But why aren't the integration boundaries -a and a, as they are in the case the rotation is around the x axis?

tiny-tim
Homework Helper
Hi tiny-tim,
But why aren't the integration boundaries -a and a, as they are in the case the rotation is around the x axis?
around the y-axis, you're slicing the volume into horizontal slices, and you must count each slice only once: each slice corresponds to two equal-and-opposite values of x, so you must use x only once, ie from 0 to a (or -a to 0)

(usually, we integrate wrt y, so obviously each value of y is used only once, but here he's "cheated" by using d(coshx/a) instead of y)

around the x-axis, you're slicing the volume into vertical slices, and again you must count each slice only once: but each slice corresponds to only one value of x, so you must go all the way from -a to a That makes sense. Thank you very much, tiny-tim! You've been very helpful :-)