Custom Solutions to Stewart Calculus, Integral
Answer
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!)
$$ \int _0 ^1 \int _0 ^1 xy( e^{2-x^2-y^2}-1)\,dy\,dx \\ \; \\ =\int _0 ^1 \int _0 ^1 \frac{-1}{2}x (e^{u}-1)\,du\,dx = \int _0 ^1 \frac{-1}{2}x (e^{u}-u)\,\Big]_0 ^1\,du\,dx \\ \; \\ =\int _0 ^1 \frac{-1}{2}x (e-2)\,dx = \frac{-(e-2)}{2}\,\frac{x^2}{2}\,\Big]_0 ^1 = \frac{-(e-2)}{4}$$
- accepted
- 436 views
- $15.00
Related Questions
- Prove that $\int_{-\infty}^{\infty}\frac{\cos ax}{x^4+1}dx=\frac{\pi}{2}e^{-\frac{a}{\sqrt{2}}}(\cos \frac{a}{\sqrt{2}}+\sin \frac{a}{\sqrt{2}} )$
- Find the extrema of $f(x,y)=x$ subject to the constraint $x^2+2y^2=2$
- Need help with finding equation of the plane containing the line and point. Given the symmetric equation.
- Calculus Questions
- Need help with integrals (Urgent!)
- Prove that $\int_0^1 \left| \frac{f''(x)}{f(x)} \right| dx \geq 4$, under the given conditions on $f(x)$
-
Find a general solution for the lengths of the sides of the rectangular parallelepiped with the
largest volume that can be inscribed in the following ellipsoid - Measure Theory and the Hahn Decomposition Theorem