# Thread: Triple Integral, Spherical Coordinates

1. ## Triple Integral, Spherical Coordinates

I am confused on how the bounds on the question are found. Here's the question.

Use spherical coordinates to find the mass of the solid enclosed by the hemispherical region $\displaystyle x^2+y^2+z^2<4$ and $\displaystyle 0<y$ with a density function (x,y,z)=
$\displaystyle \sqrt{1+(x^2+y^2+z^2)^\frac{3}{2})}$

also, the < signs are suppose to be "less than or equal to"

2. Hello,

The bound for y will be 0 and 2 :
0<y²<4-x²-z²<4

y>0
-> y<sqrt(4)=2

0<y<2

Are there any information about x or z ?

3. Nope, thats all the info given.

4. Originally Posted by Gotovina7
I am confused on how the bounds on the question are found. Here's the question.

Use spherical coordinates to find the mass of the solid enclosed by the hemispherical region $\displaystyle x^2+y^2+z^2<4$ and $\displaystyle 0<y$ with a density function (x,y,z)=
$\displaystyle \sqrt{1+(x^2+y^2+z^2)^\frac{3}{2})}$

also, the < signs are suppose to be "less than or equal to"
Looks to me like the hemisphere is the right half of a sphere centred on the origin.

You're told to work in spherical coordinates, right? So:

Radius: $\displaystyle 0 \leq r \leq 4$.

Azimuthal angle in the xy-plane: $\displaystyle -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.

Polar angle from the z-axis: $\displaystyle 0 \leq \phi \leq \pi$.

Of interest: Spherical Coordinates

5. Since $\displaystyle {\rho}^{2}=x^{2}+y^{2}+z^{2}$

We can sub it into the density function and get

$\displaystyle \sqrt{1+{\rho}^{3}}$

6. I am just a bit confused on how you find the "phi" and "theta" for this question. The bound is suppose to be 0 to Pi for phi and theta. By the way, that website on spherical coordinates was useful

7. I think your set up should look something like this:

$\displaystyle \int_{0}^{\pi}\int_{0}^{\pi}\int_{0}^{2}\left[{\rho}^{2}\sqrt{1+{\rho}^{3}}sin{\phi}\right]d{\rho}d{\phi}d{\theta}$

8. Originally Posted by galactus
I think your set up should look something like this:

$\displaystyle \int_{0}^{\pi}\int_{0}^{\pi}\int_{0}^{2}\left[{\rho}^{2}\sqrt{1+{\rho}^{3}}sin{\phi}\right]d{\rho}d{\phi}d{\theta}$
Yep that's it. Is theta 0 to Pi because its half a sphere? Also, why is the phi 0 to Pi also, I can't really picture it.

EDIT: Ok, I can see why its Pi now, I am confused from the question. I can't really see what [Math] x^2+y^2+z^2<4 [/tex] and $\displaystyle 0<y$ is. What do these regions mean?

9. Originally Posted by galactus
I think your set up should look something like this:

$\displaystyle \int_{0}^{\pi}\int_{0}^{\pi}\int_{0}^{2}\left[{\rho}^{2}\sqrt{1+{\rho}^{3}}sin{\phi}\right]d{\rho}d{\phi}d{\theta}$
Ahh yes. I misread my x- and y-axes. $\displaystyle 0 \leq \theta \pi$ indeed.

10. Originally Posted by Gotovina7
Yep that's it. Is theta 0 to Pi because its half a sphere? Also, why is the phi 0 to Pi also, I can't really picture it.

EDIT: Ok, I can see why its Pi now, I am confused from the question. I can't really see what [Math] x^2+y^2+z^2<4 [/tex] and $\displaystyle 0<y$ is. What do these regions mean?
You're told it's a hemisphere in the question: " ....enclosed by the hemispherical region ...."

Furthermore:

Originally Posted by mr fantastic
Looks to me like the hemisphere is the right half of a sphere centred on the origin.

[snip]
because 0 < y => y > 0.