# [SOLVED] Calculate a volume + theoretical question

• July 11th 2009, 03:51 PM
arbolis
[SOLVED] Calculate a volume + theoretical question
1)Identify the region of integration and calculate the following integral using spherical coordinates : $\int_{-3}^{3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int _0^{\sqrt{9-x^2-y^2}} z\sqrt{x^2+y^2+z^2} dzdydx$.

My attempt : I notice that the integrand is not the Jacobian of the change of variable spherical to Cartesian but almost. I mean that $z\sqrt{x^2+y^2+z^2}=\rho ^2 \cos \phi$ instead of $\rho ^2 \sin \phi$. Thus I am at a loss.
I don't know how to answer the question.

2) I'd like to know a general method (i.e. applicable in all cases) that transform a complicated solid (like a cone, a paraboloid, hyperboloid or anything different from a cylinder, cube and sphere) into a cube. I'd appreciate any book reference or an example with the above exercise.
I ask this because even though it might be difficult to find such a transformation, if I success in finding it then its Jacobian shouldn't be a problem to find and so the volume of the complicated solid should be quite easy to solve.
• July 11th 2009, 04:22 PM
Random Variable
You're integrating over a hemisphere of radius 3

$\int^{3}_{0} \int^{\pi /2}_{0} \int^{2 \pi}_{0} \rho^{4} \cos \phi \sin \phi \ d \theta \ d \phi \ d\rho$

$= 2 \pi \int^{3}_{0} \int^{\pi /2}_{0} \rho^{4} \cos \phi \sin \phi \ d \phi \ d\rho$

let $u = \sin \phi$

$= 2 \pi \int^{3}_{0} \int^{1}_{0} \rho^{4} u \ du\ d\rho$

$= \pi \int^{3}_{0} \rho^{4} \ d\rho$

$= \frac {243 \pi}{5}$
• July 11th 2009, 04:25 PM
HallsofIvy
Quote:

Originally Posted by arbolis
1)Identify the region of integration and calculate the following integral using spherical coordinates : $\int_{-3}^{3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int _0^{\sqrt{9-x^2-y^2}} z\sqrt{x^2+y^2+z^2} dzdydx$.

My attempt : I notice that the integrand is not the Jacobian of the change of variable spherical to Cartesian but almost. I mean that $z\sqrt{x^2+y^2+z^2}=\rho ^2 \cos \phi$ instead of $\rho ^2 \sin \phi$. Thus I am at a loss.

The integrand has nothing to do with the "region of integration". Looking at the "inside integral", z ranges from z= 0 to $z= \sqrt{9- x^2-y^2}$ which is the upper half of the sphere $z^2= 9- x^2- y^2$ or $x^2+ y^2+ z^2= 9$. Then, y ranges from $y= -\sqrt{9- x^2}$ to $y= \sqrt{9- x^2}$ so it covers $y= 9- x^2$, the circle $x^2+ y^2= 9$, the circle, in the xy-plane with center at (0,0) and radius 3. It should be no surprise that the limits of integration on x are -3 to 3. The region of integration is a hemisphere with center at (0,0,0) and radius 3. Yes, the integrand, in polar coordinates is $\rho^2 cos(\phi)$ so you the integral, in polar coordindates is
$\int_{\phi= 0}^{\pi/2}\int_{\theta= 0}^{2\pi}\int_{\rho= 0}^3 \rho^2 cos(\phi)d\rho d\theta d\phi$

Quote:

I don't know how to answer the question.

2) I'd like to know a general method (i.e. applicable in all cases) that transform a complicated solid (like a cone, a paraboloid, hyperboloid or anything different from a cylinder, cube and sphere) into a cube. I'd appreciate any book reference or an example with the above exercise.

I ask this because even though it might be difficult to find such a transformation, if I success in finding it then its Jacobian shouldn't be a problem to find and so the volume of the complicated solid should be quite easy to solve.
There is no good way to change a "smooth" surface, like a sphere or paraboloid, into a surface with edges like a cube. And I can't imagine that it is a good idea to do that since, in general, smooth surfaces are better than non-smooth!
• July 11th 2009, 04:29 PM
galactus
We have $z\sqrt{x^{2}+y^{2}+z^{2}}$

But $z={\rho}cos{\phi}, \;\ x={\rho}sin{\phi}cos{\theta}, \;\ y={\rho}sin{\phi}sin{\theta}$

Plugging these all in and simplifying and it whittles down to

${\rho}^{2}cos{\phi}$

Multiply by the ${\rho}^{2}sin{\phi}$ that comes along with

spherical integration and get ${\rho}^{4}sin{\phi}cos{\phi}$

$\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\int_{0}^{3 }{\rho}^{4}sin{\phi}cos{\phi}d{\rho}d{\phi}d{\thet a}$
• July 11th 2009, 05:48 PM
arbolis
Ok, thank you very much to all. I'll take time to think about the problem.
Quote:

Originally Posted by HallsofIvy
There is no good way to change a "smooth" surface, like a sphere or paraboloid, into a surface with edges like a cube. And I can't imagine that it is a good idea to do that since, in general, smooth surfaces are better than non-smooth!

Ok. And changing a smooth surface into another one (for example a sphere)?
For instance, in this thread : http://www.mathhelpforum.com/math-he...lume-cone.html, I don't know how NCA came up with such a transformation to find the volume of a cone. Precisely I don't understand how he found that by putting $x=\sqrt2 r \cos \theta$, etc. he could find the volume.

Edit: I understand the first question now. Thank you so much. HallsofIvy forgot to multiply the integrand by $\rho ^2 \sin \phi$, but otherwise everything's as clear as vacuum. So I'm done with the first question, but the second is still open.
• July 11th 2009, 06:39 PM
Random Variable
Quote:

Originally Posted by arbolis
Ok, thank you very much to all. I'll take time to think about the problem.
And about Ok. And changing a smooth surface into another one (for example a sphere)?
For instance, in this thread : http://www.mathhelpforum.com/math-he...lume-cone.html, I don't know how NCA came up with such a transformation to find the volume of a cone. Precisely I don't understand how he found that by putting $x=\sqrt2 r \cos \theta$, etc. he could find the volume.

It is easier to integrate over a circle than over an ellipse.

If you had $\frac {x^{2}}{a} + \frac {y^{2}}{b} \le c$, then you would let $x = \sqrt{a} r \cos \theta$ and $y = \sqrt{b} r \sin \theta$
• July 11th 2009, 06:48 PM
arbolis
Quote:

Originally Posted by Random Variable
It is easier to integrate over a circle than over an ellipse.

If you had $\frac {x^{2}}{a} + \frac {y^{2}}{b} \le c$, then you would let $x = \sqrt{a} r \cos \theta$ and $y = \sqrt{b} r \sin \theta$

Ah!!! I understand now. Finally I'm done with this thread. Thanks a million to all!