# Yet another triple integral..

• Jun 4th 2010, 11:42 PM
Em Yeu Anh
Yet another triple integral..
This is certainly not my favorite calculus topic.

$\int_{-7}^7\int_{-\sqrt{49-y^2}}^{\sqrt{49-y^2}}\int_{-\sqrt{49-x^2-y^2}}^{\sqrt{49-x^2-y^2}} (x^2z+y^2z+z^3)dzdxdy$

I am to solve this by converting to spherical coordinates.
I guessed the region of integration was the volume of the sphere $x^2+y^2+z^2=49$.

$\int_0^{2\pi}\int_0^{\pi}\int_0^7{\rho}^5cos{\phi} sin{\phi}d{\rho}d{\theta}d{\phi}$

• Jun 4th 2010, 11:53 PM
matheagle
Quote:

Originally Posted by Em Yeu Anh
This is certainly not my favorite calculus topic.

$\int_{-7}^7\int_{-\sqrt{49-y^2}}^{\sqrt{49-y^2}}\int_{-\sqrt{49-x^2-y^2}}^{\sqrt{49-x^2-y^2}} (x^2z+y^2z+z^3)dzdxdy$

I am to solve this by converting to spherical coordinates.
I guessed the region of integration was the volume of the sphere $x^2+y^2+z^2=49$.

$\int_0^{2\pi}\int_0^{\pi}\int_0^7{\rho}^5cos{\phi} sin{\phi}d{\rho}d{\theta}d{\phi}$

The bounds are correct and the answer can be zero.
The integrand is

$x^2z+y^2z+z^3=z(x^2+y^2+z^2)=(\rho\cos\phi)(\rho^2 )$

Multipling that by the jacobian and then breaking this into 3 integrals we have

$\int_0^{2\pi}d{\theta}\int_0^{\pi}\cos{\phi}\sin{\ phi}d{\phi}\int_0^7{\rho}^5 d{\rho}=0$

since $\int_0^{\pi}\cos{\phi}\sin{\phi}d{\phi}=0$
• Jun 5th 2010, 12:03 AM
Em Yeu Anh
Quote:

Originally Posted by matheagle
The bounds are correct and the answer can be zero.
The integrand is

$x^2z+y^2z+z^3=z(x^2+y^2+z^2)=(\rho\cos\phi)(\rho^2 )$

Multipling that by the jacobian and then breaking this into 3 integrals we have

$\int_0^{2\pi}d{\theta}\int_0^{\pi}\cos{\phi}\sin{\ phi}d{\phi}\int_0^7{\rho}^5 d{\rho}=0$

since $\int_0^{\pi}\cos{\phi}\sin{\phi}d{\phi}=0$

Oh good! I was at least on the right track for this one, an improvement..
Would it be wrong to take:
$2\int_0^{2\pi}\int_0^{\pi/2}\int_0^7{\rho}^5cos{\phi}sin{\phi}d{\rho}d{\thet a}d{\phi}$
To obtain a positive volume for this?

• Jun 5th 2010, 12:07 AM
matheagle
NO, that's not correct. You need symmetry in BOTH the region and the integrand
This isn't volume, that's the integral of ONE.

$1> 0$ hence $\int\int\int 1dV \ge \int\int\int 0dV =0$

BUT your integrand isn't strictly positive $z\rho^2$ can be both positive and negative and it's balanced in the sense that the positive and negative parts of z cancel each other off to give you the answer of ZERO.