# A simple calculation that got me stumped.

1. Jun 22, 2012

### MathematicalPhysicist

I need to grade some work, and there's an integral to calcaulte:

compute the area of y^2+z^2=2ax where x is between a and y^2/a.

I tried using the coordinates:

$y=\sqrt{2as}\cos(t) \ z=\sqrt{2as}\sin(t) \ x=s$

But I don't see how to find the limits of integration with respect to s and t.

Thanks.

The higher you get in maths, the less you remember what you learned in earlier years. :-(

2. Jun 22, 2012

### Millennial

I rewrite it as $x^2+y^2=2az$ where z is between a and $x^2/a$. I just renamed the variables: y as x, z as y and x as z (I am more comfortable that way.)

Written explicitly as a function of x, y is $y=\sqrt{2az-x^2}$. We know the limits, and hence this integral becomes
$$\int_{a}^{x^2/a}\sqrt{2az-x^2}dx$$

I think you can go on from there.

3. Jun 22, 2012

### mathwonk

solving your own equations says that a ≤ s = x ≤ y^2/a = 2s cos^2(t). That gives s ≥ a.

Then canceling s, from s ≤ 2s cos^2(t), we have

1 ≤ 2cos^2(t), so cos^2(t) ≥ 1/2, so -π/4 ≤ t ≤ π/4.

if that open ended inequality for s bothers you, note that your surface is unbounded in the x direction.

I.e. for any x ≥ a >0, it is easy to solve for y such that 2ax ≥ y^2 ≥ ax, and then since 2ax - y^2 ≥ 0,

one can also solve for z such that z^2 = 2ax - y^2, or z^2 + y^2 = 2ax.

Last edited: Jun 22, 2012
4. Jun 22, 2012

### Millennial

Same with me. Your solution puzzles me. I think there is something I am missing in the question.

5. Jun 22, 2012

### mathwonk

I changed a mistake in mine. (I canceled s/s and got 0.)

in your integral what does it mean to let x range from a to x^2/a?

6. Jun 22, 2012

### viraltux

Hi MP,

How about this; if $x$ was from 0 to $\frac{y^2}{a}$ then we could use polar coordinates to integrate using the radius value based on θ, so using Pythagoras we have that $cos θ = \frac{y^2}{2ay^{\frac{2}{a}}}$ and therefore $y = (2a\cosθ) ^{\frac{1}{2-\frac{2}{a}}}$ so the squared radius will be

$$r(θ)^2 = 2a((2a\cosθ) ^{\frac{1}{2-\frac{2}{a}}})^{\frac{2}{a}}$$

Now we simply polar integrate it from 0 to 2π $\frac{1}{2}\int_0^{2π} r(θ)^2 dθ$ but since $x$ is not from 0 to $\frac{y^2}{a}$ but from $a$ to $\frac{y^2}{a}$ we need to take away that part, so finally we have:

$$\frac{1}{2}\int_0^{2π} r(θ)^2 dθ - \frac{1}{2}\int_0^{2π}\min( r(θ)^2 , 2a^2) dθ$$

I didn't thoroughly check it but I think you get the idea.

Last edited: Jun 22, 2012