# Triple Integral

• April 23rd 2009, 11:55 AM
Showcase_22
Triple Integral
Quote:

Let $\Omega$ be the three dimensional region defined by $\Omega:=\{(x,y,z)| z\geq 0, x^2+y^2 \leq z^2 \leq 1-x^2-y^2 \}$.

a). Sketch the region $\Omega$
b). Evaluate $\int \int \int_{\Omega} dx \ dy \ dz$

Hint: It may be helpful to use spherical polar coordinates.
a). For this part I got an ice cream cone shape. I tried drawing it on Matlab, but I kept getting red error messages!

This ice cream cone has $0 \leq z \leq 1$ and $x^2+y^2= \left( \frac{1}{\sqrt{2}} \right)^2$.

I found this helpful for part b).!

b). Is the required integral $2 \int_0^{\frac{1}{\sqrt{2}}} \int_0^z \int_{0}^y \ z \ dx \ dy \ dz+ \frac{1}{2} \int_0^{\frac{\pi}{2}} \int_0^{2 \pi} \int_0^{\frac{1}{\sqrt{2}}} \ r^3 sin (\phi) \ dr \ d \theta \ d \phi$?

Also, is there an easier way to write this integral???
• April 23rd 2009, 12:36 PM
running-gag
Quote:

Originally Posted by Showcase_22
a). For this part I got an ice cream cone shape. I tried drawing it on Matlab, but I kept getting red error messages!

This ice cream cone has $0 \leq z \leq 1$ and $x^2+y^2= \left( \frac{1}{\sqrt{2}} \right)^2$.

I found this helpful for part b).!

b). Is the required integral $2 \int_0^{\frac{1}{\sqrt{2}}} \int_0^z \int_{0}^y \ z \ dx \ dy \ dz+ \frac{1}{2} \int_0^{\frac{\pi}{2}} \int_0^{2 \pi} \int_0^{\frac{1}{\sqrt{2}}} \ r^3 sin (\phi) \ dr \ d \theta \ d \phi$?

Also, is there an easier way to write this integral???

Hi

OK for the ice cream (Happy)

Concerning the volume, I get something more simple than you

$\int_0^{\frac{\pi}{4}} \int_0^{2 \pi} \int_0^{1} \ r^2 sin (\phi) \ dr \ d \theta \ d \phi$

Anyway your calculation cannot be true because you are required to find the volume and integrating $z \ dx \ dy \ dz$ and $r^3 sin (\phi) \ dr \ d \theta \ d \phi$ do not lead to a volume
• April 23rd 2009, 12:43 PM
Showcase_22
Quote:

$r^3 sin (\phi) \ dr \ d \theta \ d \phi$ do not lead to a volume
This is probably a bit weird, but why...?

I see what you've done for the integral and it is a lot better than what i've done. (Doh)
• April 23rd 2009, 12:52 PM
running-gag
Quote:

Originally Posted by Showcase_22
This is probably a bit weird, but why...?

You can see it as a matter of homogeneity (units)

Volume are expressed by integrating :
Cartesian : dx dy dz (which are all in meters, therefore volume is in cube meter)
Cylindrical : dr r dtheta dz (dr, r and dz are in meters)
Spherical : r² sin(phi) dr dtheta dphi (r and dr are in meters)