Custom Solutions to Stewart Calculus, Integral
Answer
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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!)
$$ \int _0 ^1 \int _0 ^1 xy( e^{2-x^2-y^2}-1)\,dy\,dx \\ \; \\ =\int _0 ^1 \int _0 ^1 \frac{-1}{2}x (e^{u}-1)\,du\,dx = \int _0 ^1 \frac{-1}{2}x (e^{u}-u)\,\Big]_0 ^1\,du\,dx \\ \; \\ =\int _0 ^1 \frac{-1}{2}x (e-2)\,dx = \frac{-(e-2)}{2}\,\frac{x^2}{2}\,\Big]_0 ^1 = \frac{-(e-2)}{4}$$
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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