# Polar Coordinates

1. Aug 28, 2014

### Zondrina

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

My answer seems to differ from the books answer, so I'm wondering where something has gone wrong.

Find the volume inside the prism bounded by the planes $y = x$, $y = 0$, $x = \frac{a}{\sqrt{2}}$, $z = 0$ and the cone $az = h\sqrt{x^2 + y^2}$.

2. Relevant equations

We want the area under $z = \frac{h}{a} \sqrt{x^2 + y^2}$

3. The attempt at a solution

When i visualized this, I saw a triangle in the x-y plane formed by $y = x$, $y = 0$, and $x = \frac{a}{\sqrt{2}}$, which gets cut by $z = \frac{h}{a} \sqrt{x^2 + y^2}$ in the x-y-z plane.

So we want:

$$V_R = \int \int_R \frac{h}{a} \sqrt{x^2 + y^2} dA$$

Using polar co-ordinates gives:

$(1) rsin(\theta) = 0 \quad$
$(2) rsin(\theta) = rcos(\theta)$
$(3) rcos(\theta) = \frac{a}{\sqrt{2}}$
$(4) f(rcos(\theta), rsin(\theta)) = \frac{h}{a} r$

Looking at $(2)$ implies $\theta = \frac{\pi}{4}$ so that $0 ≤ \theta ≤ \frac{\pi}{4}$.

The limits for $r$ have proven a little tricky. The answer is listed as:

$\frac{a^2 h}{12} [1 + \frac{3 \sqrt{2}}{2} \log(1 + \sqrt{2})]$

I have obtained the answer $\frac{a^2 h \pi}{12}$ using $0 ≤ r ≤ a$, which is obviously incorrect, but indicates my limits for $\theta$ make physical sense, so I believe the limits for $r$ are the issue. I can't seem to visualize them properly right now.

2. Aug 28, 2014

### Staff: Mentor

3. Aug 28, 2014

### Zondrina

Calculus is the bread and butter of all that is true. So forgive me if I'm firm about using it.

Where is the cylinder coming from exactly? The lines $y=0$, $y=x$ and $x = \frac{a}{\sqrt{2}}$ give a triangle with a varying height in the x-y plane (based on $\frac{a}{\sqrt{2}}$).

4. Aug 28, 2014

### Staff: Mentor

I thought I saw a cylinder with cone defining its upper edge but had forgotten about the x=a/sqrt(2) plane

5. Aug 28, 2014

### vela

Staff Emeritus
Why are you even using polar coordinates here? It's more straightforward to describe the area of integration using cartesian coordinates.

6. Aug 28, 2014

### Zondrina

Re-arranging the equation for the cone a bit:

$az = h \sqrt{x^2 + y^2} \Rightarrow \frac{z^2}{h^2} = \frac{x^2}{a^2} + \frac{y^2}{a^2}$

I see that it opens up in the z-direction and the top of the cone should be a perfect circle of radius $a$. I still can't see how to get the radius otherwise.

The book says it is possible to find the solution with polar co-ordinates, and I would like to find a solution using them.

7. Aug 28, 2014

### LCKurtz

You have that triangle you described in the xy plane. Use it to get the polar limits. $r$ goes from $0$ to the $r$ value on the line $x=\frac a{\sqrt 2}$. What is the equation of that line in polar coordinates?

Last edited: Aug 28, 2014
8. Aug 28, 2014

### Zondrina

That would correspond to equation $(3)$ in my first post I believe.

$$r \cos(\theta) = \frac{a}{\sqrt{2}}$$

Though I doubt that $0 ≤ r ≤ \frac{a}{cos(\theta) \sqrt{2}}$

9. Aug 28, 2014

### LCKurtz

Why do you doubt it? If you start at $r=0$ and move through the region in the $r$ direction, isn't that the line you hit? And I would suggest expressing that with $\sec\theta$ when you set up the integral.

10. Aug 28, 2014

### Zondrina

I am never going to believe wolfram again.

I think I understand now that you have clarified that. Thank you.