# Thread: [SOLVED] Calculate a volume + theoretical question

1. ## [SOLVED] Calculate a volume + theoretical question

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

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

2. You're integrating over a hemisphere of radius 3

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

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

let $\displaystyle u = \sin \phi$

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

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

$\displaystyle = \frac {243 \pi}{5}$

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

My attempt : I notice that the integrand is not the Jacobian of the change of variable spherical to Cartesian but almost. I mean that $\displaystyle z\sqrt{x^2+y^2+z^2}=\rho ^2 \cos \phi$ instead of $\displaystyle \rho ^2 \sin \phi$. 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 $\displaystyle z= \sqrt{9- x^2-y^2}$ which is the upper half of the sphere $\displaystyle z^2= 9- x^2- y^2$ or $\displaystyle x^2+ y^2+ z^2= 9$. Then, y ranges from $\displaystyle y= -\sqrt{9- x^2}$ to $\displaystyle y= \sqrt{9- x^2}$ so it covers $\displaystyle y= 9- x^2$, the circle $\displaystyle x^2+ y^2= 9$, 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 $\displaystyle \rho^2 cos(\phi)$ so you the integral, in polar coordindates is
$\displaystyle \int_{\phi= 0}^{\pi/2}\int_{\theta= 0}^{2\pi}\int_{\rho= 0}^3 \rho^2 cos(\phi)d\rho d\theta d\phi$

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!

4. We have $\displaystyle z\sqrt{x^{2}+y^{2}+z^{2}}$

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

Plugging these all in and simplifying and it whittles down to

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

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

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

$\displaystyle \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{2}}\int_{0}^{3 }{\rho}^{4}sin{\phi}cos{\phi}d{\rho}d{\phi}d{\thet a}$

5. Ok, thank you very much to all. I'll take time to think about the problem.
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 $\displaystyle x=\sqrt2 r \cos \theta$, etc. he could find the volume.

Edit: I understand the first question now. Thank you so much. HallsofIvy forgot to multiply the integrand by $\displaystyle \rho ^2 \sin \phi$, but otherwise everything's as clear as vacuum. So I'm done with the first question, but the second is still open.

6. 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 $\displaystyle x=\sqrt2 r \cos \theta$, etc. he could find the volume.
It is easier to integrate over a circle than over an ellipse.

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

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

If you had $\displaystyle \frac {x^{2}}{a} + \frac {y^{2}}{b} \le c$, then you would let $\displaystyle x = \sqrt{a} r \cos \theta$ and $\displaystyle y = \sqrt{b} r \sin \theta$
Ah!!! I understand now. Finally I'm done with this thread. Thanks a million to all!