# Thread: Converting from Rectangular to Spherical Coordinates

1. ## Converting from Rectangular to Spherical Coordinates

So I'm asked to evaluate the integral by changing to spherical coordinates, $\displaystyle \displaystyle \int_{0}^{1} \int_{0}^{\sqrt{1-x^{2}}} \int_{\sqrt{x^{2} + y^{2}}}^{\sqrt{2 - x^{2} - y^{2}}}xy \, dz \, dy \, dx$.

I see how do to do all the plugging in, but I'm not sure how to determine the limits of integration. I know my radius is going to start at 0, but I want it to be limited by some function of $\displaystyle \phi$. I can see that I want to be looking only at what's above the xy-plane and toward the positive side of x, so I want $\displaystyle -\pi/2 \leq \theta \leq \pi/2$ and $\displaystyle \pi/2 \leq \phi \leq \pi$. So I guess all that I don't see is what function bounds the radius on top.

2. ... Er, I think what I really want is $\displaystyle 0 \leq \phi \leq \pi/2$, but anyway.

3. Originally Posted by ragnar
... Er, I think what I really want is $\displaystyle 0 \leq \phi \leq \pi/2$, but anyway.
No, did you sketch the region? you're integrating over the volume of part of an ice cream cone, the bottom part comes from the cone $\displaystyle \displaystyle z^2 = x^2 + y^2$, the top part comes from the sphere $\displaystyle x^2 + y^2 + z^2 = 2$. You're in the first quadrant only.

You have $\displaystyle \displaystyle 0 \le \theta \le \frac {\pi}2$

$\displaystyle \displaystyle 0 \le \phi \le \frac {\pi}4$, and

$\displaystyle \displaystyle 0 \le \rho \le \sqrt 2$

Draw the figure and see if you can figure these out. begin with the traces

4. Oh, right, my x is positive and y positive and z positive, so I'm just looking at the first octant, hence $\displaystyle 0 \leq \theta \leq \pi/2$. Because it's a cone and when keeping y fixed $\displaystyle x^{2} = z^{2}$ so it's a 45-degree angle hence $\displaystyle 0 \leq \phi \leq \pi/4$. And because the top part is a portion of a circle the radius reaches out to a constant end-point, in this case $\displaystyle \sqrt{2}$ when we plug in $\displaystyle x = y = 0$ for the upper limit. Hopefully that's the right way to think of it.

5. Originally Posted by ragnar
Oh, right, my x is positive and y positive and z positive, so I'm just looking at the first octant, hence $\displaystyle 0 \leq \theta \leq \pi/2$.
yup

Because it's a cone and when keeping y fixed $\displaystyle x^{2} = z^{2}$ so it's a 45-degree angle hence $\displaystyle 0 \leq \phi \leq \pi/4$.
you mean if you put y = 0 you get that. I suppose you can think of it that way, but it seems somewhat limited. Notice that if you view the figure sideways (positive z-axis pointing up, the xy-plane is the horizontal), a part of the figure will be a right triangle whose height and base lengths are z. this is an isosceles right triangle, and hence the other two angles are $\displaystyle \frac {\pi}4$, which gives you your $\displaystyle \phi$

And because the top part is a portion of a circle the radius reaches out to a constant end-point, in this case $\displaystyle \sqrt{2}$ when we plug in $\displaystyle x = y = 0$ for the upper limit. Hopefully that's the right way to think of it.
$\displaystyle \rho$ is the distance from the origin to the outer shell of the solid figure, which in this case is the distance from the origin to the sphere. the distance is therefore the radius of the sphere, which is $\displaystyle \sqrt 2$, and that gives you your $\displaystyle \rho$

from these you can find the limits. the moral? if possible, draw the region they are talking about, usually you can use basic geometry/trigonometry to figure out the limits.