Set up and evaluate the definite integral for the area of the surface formed by revolving the graph of y=9-x^2 about the y-axis

Question

anonymous · Answer

I know the answer is E, but how do I get that answer?

tkhunny · Answer

Please show the standard, general formulation for this sort of thing.  Let's see if you're on the right page.

anonymous · Answer

Ok hold on...

anonymous · Answer

$$2\pi \int\limits_{-3}^{3}\sqrt{1+(-2y)^2} dy$$ Is this the correct equation?

anonymous · Answer

Well sorta?

anonymous · Answer

The -2y; I got that from getting the derivative of 9-y^2

tkhunny · Answer

Two things funny about that.

1) Where is the missing 'x' for the radius of the pieces you are spinning?
2) Are you sure you want [-3,3] rotated through 2$\pi$?  You may be counting things twice.

What say you?

It may help if you answer my question directly.  The GENERAL way to proceed is:

About the y-axis $\int\limits_{a}^{b}2\pi x\sqrt{1+{\left(\dfrac{dx}{dy}\right)^{2}}}\;dy$

anonymous · Answer

Got it! 2pi intergal min 0 max 3 and place the x in front of the square root. Everything inside the sqrt is correct

anonymous · Answer

Thank you! I tried 2pi intergal min 0 max 3 before but this whole time I was missing the x in front of the sqrt

tkhunny · Answer

If $y = 9-x^{2}\;then\;\dfrac{dy}{dx} = -2x$, but we don't need that.

We need $\dfrac{dx}{dy} = \dfrac{1}{-2x}$