# Homework Help: Why is the limit for theta = pi/2 instead of 2pi?

1. Dec 16, 2015

### qq545282501

1. The problem statement, all variables and given/known data
use cylindrical coordinates to find the volume of the solid which is under z=xy, above xy-plane and inside the cylinder x^2+y^2=2x

2. Relevant equations

3. The attempt at a solution
$$\int_{0}^{pi/2} \int_{0}^{2cos\theta} \int_{0}^{r^2\cos\theta\sin\theta} r\, dz \, dr \, d\theta$$

my professor had it pi/2, but when I do it myself, I dont see how its pi/2. 2pi would make more sense to me since it has a cylinder base.

any help is appreciated

2. Dec 16, 2015

### blue_leaf77

The curve $r = 2\cos \theta$ is fully circulated when $\theta$ goes from 0 to $\pi$, so I would actually expect the limit to be from 0 to $\pi$.

3. Dec 16, 2015

### qq545282501

thank you ! Oh i see why its 0 to pi, i just graphed that curve, i see its not centered at (0,0), its in the first and fourth quadrant.
so i must have made a mistake when I took my notes in the class then?

4. Dec 16, 2015

### blue_leaf77

Either ask one of your colleagues or confirm directly to the lecturer.

5. Dec 16, 2015

### qq545282501

no more classes left for this course though :(

anyway,I just realized that if its 0 to pi, the volume become 0

6. Dec 16, 2015

### blue_leaf77

Ah I see why it should indeed be from 0 to $\pi/2$. The cylinder occupies $x>0$ region of space, but in this region the surface $z=xy$ is positive if $y>0$ and negative if $y<0$. Because the volume is required to be above xy plane where z is positive, the part of $z=xy$ surface which is negative (that with $y<0$ and $x>0$) must be ruled out.

7. Dec 16, 2015

### qq545282501

oh interesting, I understand now. in order for z to be positive, y must be positive, which means first quadrant only.
thx a lot !

8. Dec 16, 2015

### blue_leaf77

You are welcome.

9. Dec 16, 2015

### Ray Vickson

The third quadrant $x<0$ and $y<0$ would also be allowed by $0 \leq z \leq xy$, but the cylindrical base $x^2 + y^2 = 2x$ lies completely in the first and fourth quadrants. Altogether, when you impose both conditions your region is restricted to the first quadrant.

10. Dec 16, 2015

### blue_leaf77

I was actually aware of that, that was infact what enables me to write post #2.

11. Dec 16, 2015

### Ray Vickson

Sorry: wrong response.

I had meant to respond to the OP's post #7, which mistakenly stated that to have $z > 0$ it was necessary to have $y > 0$.