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i am given the following problem: $$\iiint_K e^z$$ where :

$$ K \{x^2+y^2+z^2 \leq 1\} $$

here are the steps that i did= $$z=\sqrt{1-x^2-y^2}$$

then the problem becomes a double integral problem : $$\iiint_K e^z = \iint_D e^{\sqrt{1-x^2-y^2}} dxdy$$

using the polar coordinates we get: $$x=r \cos(t)$$ $$y=r\sin(t)$$ $$0\leq t \leq 2\pi$$ $$0\leq x\leq1$$ partial derivitives of $z_x, z_y$ are given by: $$z_x= \frac{-x}{\sqrt{1-x^2-y^2}}$$
$$z_y= \frac{-y}{\sqrt{1-x^2-y^2}}$$ now the integral becomes: $$\int_0^1\int_0^{2\pi}\frac{e^\sqrt{1-r^2} }{\sqrt{1-r^2}}r dtdr $$ $$=2\pi \int_0^1 e^udu $$ $$=2\pi(e-1)$$ my question here if this operation is mathematically legal, given that this is really a projection on the $xy$ plane right?

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No, it is not legitimate to project the surface of a sphere onto a disk and integrate it there unless you also include a factor to account for the ratio between a region on the surface of the sphere and the region of the disk it maps to. But even worse, you started with a volume integral and then proceeded to ignore the entire interior of the sphere and the surface of the lower hemisphere.

It seems to be time to start over from the beginning. You might consider a disk method of integration instead.

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  • $\begingroup$ would you mind providing an answer that doesnt include using the spherical coordinates? $\endgroup$
    – Reddevil
    May 31 '17 at 1:55
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For any $z_0\in[-1,1]$, the section $z=z_0$ of the ball $x^2+y^2+z^2\leq 1$ is a circle with radius $\sqrt{1-z_0^2}$, hence area $\pi(1-z_0^2)$. By Cavalieri's principle

$$ \iiint_{B}e^{z}\,dx\,dy\,dz = \int_{-1}^{1}e^{z_0}\pi(1-z_0^2)\,dz_0 =\color{red}{\frac{4\pi}{e}}$$ nice and easy.

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