Tripple Integral Homework: Finding \int\int\int_D dV

[sandy.bridge]

Homework Statement

find the value of $$\int\int\int_D{e^{(x^2+y^2+z^2)^{3/2}/2}}dV$$ given $$1\leq{x^2+y^2+z^2}\leq{3}, z^2\geq{2}(x^2+y^2), 2x\leq{y}\leq3x$$
I am reviewing tripple integrals and am having a bit of difficulty determining the limits for each part. I have,
$$arctan(2)\leq\theta\leq{arctan(3)}, 1\leq\rho\leq{\sqrt{3}}$$

but I can't seem to visualize what is happening with
$$\phi$$
Any suggestions?

 

[Dick]

Can you visualize what the surface z^2=2*(x^2+y^2) looks like?

 

[sandy.bridge]

Kind of; paraboloid? If it is indeed a paraboloid, how does one determine the angle phi when the parabloid has a curved surface?
$$\phi{_1}\leq\phi\leq\pi/2$$

 

[Dick]

Kind of; paraboloid? If it is indeed a paraboloid, how does one determine the angle phi when the parabloid has a curved surface?
$$\phi{_1}\leq\phi\leq\pi/2$$

z=2*(x^2+y^2) is a paraboloid. z^2=2*(x^2+y^2) isn't. Think a little harder. I'll give you a hint. Sketch z^2=2*x^2 in the x-z plane.

 

[sandy.bridge]

It's a cone, I believe. Now, this cone is going to intersect the outer edge of the sphere, and therefore, plugging the equation for the cone into x^2+y^2+z^2=3, we get
$$3x^2+3y^2=3\rightarrow{x^2+y^2=1}$$
Do I have the right idea so far?

 

[Dick]

It's a cone, I believe. Now, this cone is going to intersect the outer edge of the sphere, and therefore, plugging the equation for the cone into x^2+y^2+z^2=3, we get
$$3x^2+3y^2=3\rightarrow{x^2+y^2=1}$$
Do I have the right idea so far?

You have the right idea that it's a cone. Doesn't the cone determine the range of ϕ? Isn't that what you were asking about?

 

[sandy.bridge]

Indeed, it does. However, is this not limited by the point that the cone intersects the outer edge of shell? This is why I combined the equations
$$x^2+y^2+z^2=3, z^2=2(x^2+y^2)$$
This point of interesction can then be used to determine the angle from the xy-plane, which phi will merely be 90 less the angle measured.

 

[sandy.bridge]

Here is the drawing:

So $$\phi$$ will go from the angle that it intersects to pi/2

 

[Dick]

Indeed, it does. However, is this not limited by the point that the cone intersects the outer edge of shell? This is why I combined the equations
$$x^2+y^2+z^2=3, z^2=2(x^2+y^2)$$
This point of interesction can then be used to determine the angle from the xy-plane, which phi will merely be 90 less the angle measured.

Right, but I don't think there is any need to intersect z^2=2(x^2+y^2) with anything. ANY point on the cone that's above the x-y plane has the same value of phi.

 

[sandy.bridge]

This is true. I got the right answer. Thanks