# Thread: Area of a Surface of Revolution

1. ## Area of a Surface of Revolution

x = 1 + 2y^2, 1 <= y <= 2

so I figure out dy/dx = 4y

2PI integral[ (1+2y^2)*sqrt(1+16y^2)]

At this point, I've tried so many different integration techniques but I can't find the antiderivative.

2. Originally Posted by s3n4te
x = 1 + 2y^2, 1 <= y <= 2

so I figure out dy/dx = 4y

2PI integral[ (1+2y^2)*sqrt(1+16y^2)]

At this point, I've tried so many different integration techniques but I can't find the antiderivative.
I assume the rotation is around the y-axis ....?

Try the substitution $y = \frac{1}{4} \sinh u$.

3. sorry, it's around the x-axis, i thought it was y until i re-read the question. LOL, it's really simple when it's around the x-axis doh.