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- Thread starter b0t2
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- Thread starter
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- #2

- Mar 5, 2012

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Welcome to MHB, b0t2!

Let's rewrite your statements in $\LaTeX$.

Apparently your problem has:

$$x^2=4y-y^2; \quad x^2=8y-y^2; \quad y=x; \quad x=0$$

So in polar:

$$r=4\sin \phi; \quad r=8\sin \phi; \quad \phi=\pi/4; \quad \phi=\pi/2$$

The integral:

$$\int_{\pi/4}^{\pi/2} d \phi \int_{4 \sin \phi}^{8 \sin \phi} r dr$$

Your answer is $3\pi - 1/4$ which appears not to be true.

Perhaps you can clarify?

From your first statement I deduce x=y=0, but obviously that is not what you intended.

I suspect you're talking about some kind of intersection of surfaces, but I prefer not to guess.

Do you perhaps have the complete problem statement?