Change of Variables multiple integrals

[missavvy]

Homework Statement

Find the volume of the cone bounded below by z=2root(x²+y²) and above by x² + y² + z² = 1

Homework Equations

The Attempt at a Solution

Ok I have the solution, I just don't understand how to get it!

So I know I have to change into spherical coordinates but I'm having trouble with this...
so the coordinates are of the form (r, $$\varphi$$, $$\vartheta$$)

THe equation of the cone is: tan$$\varphi$$ = 1/2, r=1 is the equation of the sphere...


My question is how to I swtich from cartesian to spherical?

I thought (x,y,z) = (r, $$\varphi$$, $$\vartheta$$) = (rsin$$\varphi$$cos$$\vartheta$$, rsin$$\varphi$$sin$$\vartheta$$, rcos$$\varphi$$)

I tried setting r=1=x² + y² + z² ...
Er.. what do I "solve" for first?

Also I know it can be solved using cylindrical coordinates.
Is there one "easier" than the other?
I'm really frustrated.. sorry if this is sloppy :(

 

[vela]

You already converted to spherical coordinates when you wrote the equations for the sphere and the cone. You just need to translate those into the proper limits on the integral.

$$V = \int dv = \int_{r_0}^{r_1} \int_{\phi_0}^{\phi_1} \int_{\theta_0}^{\theta_1} r^2\sin \phi\,d\theta\,d\phi\,dr$$

A sketch may help you visualize what these limits should be.

 

[HallsofIvy]

Did you give much thought to this. In spherical coordinates,
$$x= \rho cos(\theta)sin(\phi)$$
$$y= \rho sin(\theta)sin(\phi)$$
$$z= \rho cos(\phi)$$

So that
$$2\sqrt{x^2+ y^2}= 2\sqrt{\rho^2 sin^(\phi)(cos^2(\theta)+ sin^2(\theta))}$$
$$= 2\rho sin(\phi)$$

$$z= \rho cos(\phi)$$
so "$z= 2\sqrt{x^2+ y^2}$" is
$$\rho cos(\phi)= 2\rho sin(\phi)$$
Now, what does that tell you about the limits on $\phi$?

Similarly, the equation of the sphere, $x^2+ y^2+ z^2= 1$ reduces to
$\rho^2= 1$ or, because $\rho$ is non-negative, $\rho= 1$.
That gives you the limits on $\rho$.

And the fact this region is symmetric about the z-axis gives the limits on $\theta$.

 

[missavvy]

ok .. so I get up to his point: $$\int$$ (FROM 0 to 1) (2$$\pi$$r²(-cos$$\varphi$$)) (from 0,tan^-1(1/2) ) dr

(I may have used different variables than what you suggested, but basically I get that)
How can I sub in the tan into the phi ? do I evaluate arctan(1/2) then put it into the CosPhi? it`s a messy number if I do..

 

[spamiam]

Draw a triangle with such that one of its angles has tangent of 1/2. Using the Pythagorean theorem, you should be able to find cosine of that angle.