Question

Triple Integral Problem: Convert to Spherical coordinates to evaluate:

Answer 1

\[\int\limits_{-1}^{1}\int\limits_{0}^{\sqrt{1-x ^{2}}}\int\limits_{0}^{\sqrt{1-x^{2}-y ^{2}}}e ^({-x ^{2}+y^{2}+z^{2}})^{3/2} dz dy dx\]

Answer 2

Can anyone guide me on how to convert from rectangular to spherical? That's my biggest problem the integral wont be to rough. Once I know the new limits of integration and the stuff inside the function is converted.

Answer 3

\[\int_{-1}^1\int_0^\sqrt{1-x^2}\int_0^\sqrt{1-x^2-y^2}\left[e^{-x^2+y^2+z^2}\right]^{3/2}~dz~dy~dx~~?\]
or
\[\int_{-1}^1\int_0^\sqrt{1-x^2}\int_0^\sqrt{1-x^2-y^2}e^{-\left[x^2+y^2+z^2\right]^{3/2}}~dz~dy~dx~~?\]

Answer 4

Sorry about that the second one.

Answer 5

For the integrand, the conversion is simple enough:
\[\begin{cases}
x=\rho\sin\phi\cos\theta\\
y=\rho\sin\phi\sin\theta\\
z=\rho\cos\phi\\
x^2+y^2+z^2=\rho^2\\
dx~dy~dz=\rho^2\sin\phi~d\rho~d\theta~d\phi
\end{cases}\]
The cartesian coordinates give you the hemisphere centered at (0,0,0) with radius 1. In polar coordinates, you have
\[\begin{cases}
0\le\theta\le2\pi\\
0\le\phi\le\frac{\pi}{2}\\
0\le\rho\le1
\end{cases}\]