# Triple integral

1. Nov 12, 2013

### MuIotaTau

1. The problem statement, all variables and given/known data

Integrate $f(x,y,z) = z$ over the region bounded by $z = 0$, $x^2 + 4y^2 = 4$, and $z = x + 2$,

2. Relevant equations

None.

3. The attempt at a solution

I sketched the region in question, but my drawing is so terrible that I'm afraid it'll be little help to anybody who didn't draw it themselves. To describe the region in question, it is an elliptical cylinder (i.e. cylinder whose xy-cross section is an ellipse) and it is bounded by the xy-plane on the bottom and by the plane $z = x + 2$ on the top. Given this, I set up my integral as follows:

$$I = \int_{-2}^2 \int_{-2\sqrt{1 - y^2}}^{2 \sqrt{1 - y^2}} \int_0^{x + 2} z \ dz \ dx \ dy$$

Can anybody spot a mistake so far? I've attempted the entire problem, but I would like to make sure there isn't an issue so far before I tex it all up if possible. Thanks!

2. Nov 12, 2013

### Staff: Mentor

Looks OK.

3. Nov 12, 2013

### MuIotaTau

Okay, thank you for that. So my progress is as follows:

$$I = \frac{1}{2} \int_{-2}^2 \int_{-2\sqrt{1 - y^2}}^{2\sqrt{1 - y^2}} \bigg(x^2 + 4x + 4 \bigg) \ dx \ dy$$

$$I = \frac{1}{2} \int_{-2}^2 \left[\frac{1}{3}x^3 + 2x^2 + 4x \right]_{-2\sqrt{1 - y^2}}^{2\sqrt{1 - y^2}} \ dy$$

$$I = \frac{1}{2} \int_{-2}^2 \bigg(\frac{8}{3}(1 - y^2)^{\frac{3}{2}} + 8(1 - y^2) + 8(1 - y^2)^{\frac{1}{2}} + \frac{8}{3}(1 - y^2)^{\frac{3}{2}} - 8(1 - y^2) + 8(1 - y^2)^{\frac{1}{2}} \bigg)\ dy$$

$$I = 8 \int_{-2}^{2} \bigg(\frac{1}{3}(1 - y^2)^{\frac{3}{2}} + (1 - y^2)^{\frac{1}{2}} \bigg) \ dy$$

I simply don't know how to integrate this, and I feel as if I'm missing something simple. Any suggestions?

4. Nov 12, 2013

### LCKurtz

I would try switching to polar coordinates right at this step.

5. Nov 12, 2013

### Staff: Mentor

It might be easier to switch the order of integration: dz dy dx. Using symmetry would make it a little simpler as well.
$$\frac 1 2 \int_{-2}^2 \int_0^{1/2 \sqrt{4 - x^2}} \int_0^{x + 2} z \ dz \ dy \ dx$$

If you have drawn a sketch of the region, you'll see that the volume above the upper half of the ellipse (where y ≥ 0) is the same as the volume over the lower half.

In any case, since you have invested a fair amount of time, you'd probably rather just continue from where you are. Assuming your work so far is correct, you basically have two integration problems. The first is the one to the 3/2 power. That looks like a trig substitution to me. The second looks like a trig substitution as well, but there's a shortcut. The part with the 1/2 power represents a half-circle whose radius is 1. Should be easy to get that one.

6. Nov 12, 2013

### MuIotaTau

My sleep schedule is killing me right now, so I'm gonna hit the hay for the time being, but I appreciate all these posts and I'll come back to this thread tomorrow.