# Integrating triple integral over region W

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1. Nov 12, 2016

### toforfiltum

1. The problem statement, all variables and given/known data
$$f(x,y,z)=y$$ ; W is the region bounded by the plane $x+y+z=2$, the cylinder $x^2 +z^2=1$, and $y=0$.

2. Relevant equations

3. The attempt at a solution
Since there is a plane of $y=0$, I decided that my inner integral will be $y=0$ and $y=2-x-z$. But after this I have a problem. I don't know how to proceed. How do I decide the bounds of $x$ and $z$? Can I just use the function $x^2+z^2=1$? If so, will it be $$\int_{-1}^1 \int_{-\sqrt(1-x^2)}^{\sqrt(1-x^2)} \int_ {0}^{2-x-z} y \,dy \, dz \, dx$$

Thanks.

2. Nov 12, 2016

### LCKurtz

Yes, you can do it that way. It might be easier to use polar type coordinates for the xz integral because of the circle, depending on how easy the rectangular integral is.

3. Nov 12, 2016

### toforfiltum

Yes, the rectangular integral is really horrible. It involved a lot of $cos^4 \theta$ terms.

4. Nov 12, 2016

### RUber

I agree with LCKurtz. Change to cylindrical coordinates. You have to account for the place where the plane starts to intersect the cylinder. Otherwise, you'll be integrating over the entire circle near the top, and won't get the right volume.

5. Nov 12, 2016

### toforfiltum

But I think I'm supposed to be calculating it using rectangular coordinates, because this question is in the section before the change of variables topic.

6. Nov 12, 2016

### RUber

Okay then. Maybe look at changing the order of integration? Let y be your outermost integral, since limits on x and z will both depend on how much of the circle is being cut off by the plane.
It looks like there are two parts here...the first part is just the cylinder before the plane intersects it.
The second part will be from y = 1 to y = 3. Try to define your x and z limits based on that.

7. Nov 13, 2016

### LCKurtz

I'm not sure I understand @RUber 's concerns and suggestions. I would say you definitely want to integrate in the $y$ direction first, and the remaining $x,y$ or $r,\theta$ integral will be over the whole circle in the xz plane. I think you should get $\frac{9\pi}{4}$ either way.

8. Nov 14, 2016

### toforfiltum

Ok, thanks for providing the correct answer for me to check. However, I got a wrong answer. What I would like to know is if the process of calculating the integral in rectangular coordinates really long? Because mine certainly is. I'm just wondering if I'm doing it correctly.

9. Nov 14, 2016

### toforfiltum

@LCKurtz Oh, it's fine. I've got the correct answer now, though only after a painfully long process.

Thanks!

10. Nov 14, 2016

### LCKurtz

Good job. Luckily, I didn't experience the "pain" of working it all out, since I just put your original correct triple integral into Maple. It cranked out the answer in a couple of seconds.