How do you define the limits for this triple integral?

[Inertigratus]

I had this on a test today. First everything seemed easy, but then I got stuck.
So the body over which the integral is to be taken is defined by:
1 <= x² + y² + z² <= 4
and
z >= sqrt(x² + y²)

Right now as I'm typing this I just thought that, why not plug z from the second eq. into the first? is this possible, even though it's not an equality but greater than or equal?

That would make it: (1/2) <= x² + y² <= 2
Then: 1/sqrt(2) <= z <= sqrt(2)

Is this right? If so, what about the limits for x and y, can I just change to polar coordinates and let r = radius vary from 1 to 4?

 

[spamiam]

Hi Inertigratus,

I would probably solve this problem by switching to spherical coordinates. Whenever you're trying to figure out the limits of integration, it's usually best to draw yourself a picture. What regions are described by $1 \leq x^2 + y^2 + z^2 \leq 4$ and $z \geq \sqrt{x^2 + y^2}$?

I'm not sure your method for solving for z works... For one thing, when x = 0 = y, then $z = \pm 2$ on the outer boundary.

 

[Inertigratus]

Hi Inertigratus,

I would probably solve this problem by switching to spherical coordinates. Whenever you're trying to figure out the limits of integration, it's usually best to draw yourself a picture. What regions are described by $1 \leq x^2 + y^2 + z^2 \leq 4$ and $z \geq \sqrt{x^2 + y^2}$?

I'm not sure your method for solving for z works... For one thing, when x = 0 = y, then $z = \pm 2$ on the outer boundary.

That is what I'm a little unsure of. The first inequality states that the region is between the two spheres with radius 1 and 2. The other inequality states that the height is greater than the radius... but, the radius of what?
There are two spheres with different radius...
I'm having trouble understanding the second inequality.

 

[Mute]

In spherical coordinates,

$$x = r\cos \phi \sin \theta,~y = r\sin \phi \sin \theta,~z = r\cos \theta$$

so the first inequality corresponds to $1 < r < 2$, as you said. If you plug the above relations into the second inequality, you should be able to figure out the range for the angles.

 

[Inertigratus]

sin(θ) = cos(θ) ?

 

[Inertigratus]

If I substitute z and x, y in the second inequality that's what I get. Any idea on how to get the angle?
I'm not too good at trig...

 

[spamiam]

Two ways you can figure out theta:

1) What kind of surface is described by $z = \sqrt{x^2 + y^2}$? Also if you plug in x=0 or y=0, you get the projection on the yz- or xz-plane, respectively, which will allow you to figure out the angle.

2) $\sin\theta = \cos\theta$ is equivalent to $\tan\theta = 1$. So draw yourself a right triangle with legs of equal length, and then find the measure of the angle.

 

[Mute]

sin(θ) = cos(θ) ?

Almost. Just to be clear that you're doing all the steps: Remember that you have an inequality, $z \geq \sqrt{x^2 + y^2}$. If you plug in the spherical coordinate relations, you get $r\cos\theta \geq \sqrt{r^2\cos^2\phi\sin^2\theta + r^2\sin^2\phi\sin^2\theta}$, which after some simpificatin yields

$$\cos\theta \geq \sin \theta.$$

So, your task is to find the range of $\theta$ for which this is true, knowing that the range of $\theta$ is restricted to $[0,\pi]$. If you plot both cos and sin on the same plot, you can see that this is true starting at 0 all the way to the value of theta which makes the above inequality an equality. As spamiam pointed out, $\cos\theta = \sin\theta$ is equivalent to $1 = \tan\theta$.

Also, note that this didn't put any restriction on the angle $\phi$, so it can take all values between $[0,2\pi)$.