2.4 Arc Length of a Curve and Surface Area

[STARTS AT 19:28]

PROFESSOR: Last one-- we have a function defined in terms of why again. But this time, we don't have to rearrange anything. Our function is g of y. So our surface area is the integral from 0 to 2, based on our interval. And notice, we're also still around the y-axis. So from 0 to 2, 2 pi square root of 9 minus y squared square root of 1 plus the derivative of this function. 

All right, well, this is 9 minus y squared to the 1/2. So g prime of y is 1/2 9 minus y squared to the negative 1/2 times negative 2y. And if I square that, that is going to be 1/4. Actually, that would be negative 1 y. Those two cancel out to be negative y. So we get negative y over the square root of 9 minus y squared. So if we square that, that will be y squared over 9 minus y squared. So y squared over 9 minus y squared dy, which I believe is approximately 12 pi, if my calculations are correct.

[ENDS AT 21:24]