# Custom Solutions to Stewart Calculus, Integral

I am looking for custom solution to this triple integral from Stewart Calculus (8th edition). I don't want the kind of answers available in solution manuals or given by AI assistants like chatGPT.

Sure, you came to the right place to get custom solutions given by experts.

For this integral we start by integrating with respect to $z$ : $$\int _0 ^1 \int _0 ^1 \int_0 ^{2-x^2-y^2} xy e^{z}\,dz\,dy\,dx = \int _0 ^1 \int _0 ^1 xy e^{z}\,\Big]_0 ^{2-x^2-y^2}\,dy\,dx \\ = \int _0 ^1 \int _0 ^1 xy( e^{2-x^2-y^2}-1)\,dy\,dx.$$
Then we do the $y$ integral. For this let $u=2-x^2-y^2$ to get $du=-2ydy$ (remember that in the $y$ integral we treat $x$ as a constant!)
$$\int _0 ^1 \int _0 ^1 xy( e^{2-x^2-y^2}-1)\,dy\,dx \\ \; \\ =\int _0 ^1 \int _0 ^1 \frac{-1}{2}x (e^{u}-1)\,du\,dx = \int _0 ^1 \frac{-1}{2}x (e^{u}-u)\,\Big]_0 ^1\,du\,dx \\ \; \\ =\int _0 ^1 \frac{-1}{2}x (e-2)\,dx = \frac{-(e-2)}{2}\,\frac{x^2}{2}\,\Big]_0 ^1 = \frac{-(e-2)}{4}$$

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