Double integral conversion to polar coordinates

[ProPatto16]

i have no idea how to use the functions on here to ill try my best.

$$\int$$(upper bound a lower bound 0)$$\int$$(upper bound 0 lower bound -sqrt(a²-y²) of the function x²y.dxdy

firstly trying to map it out...

i think its the quarter circle in the top left quadrant with boundaries 0 to a along y-axis and boundaries 0 to -sqrt(a²-y²) along -ve x-axis.

assuming that is right, and x=rcos$$\theta$$ amd y=rsin$$\theta$$ so then the function needing to be integrated is (r²cos²$$\theta$$*rsin$$\theta$$)rdrd$$\theta$$

the radius r is equal to a so sub in for r and integrate the function and all that. is that the right track??

sorry if its hard to understand :(

 

[tiny-tim]

Hi ProPatto16!

(have an integral: ∫ and a theta: θ and a square-root: √ )

… the radius r is equal to a so sub in for r and integrate the function and all that. is that the right track??

you were doing fine down to (r²cos²θ*rsinθ)rdrdθ …

but no, you can't sub in a for r, you have to integrate wrt r, and then put r = a (and 0)

 

[ProPatto16]

Oh. That's going to be a long integration -.-
Thank you sir!

 

[lanedance]

so do you meanbelow? (click on it to see tex code)
$$\int_0^a \int_0^{-\sqrt{y^2-a^2}}dxdy x^2y$$

 

[ProPatto16]

The second integral sign, swap those bounds around and swap a and y in the square root. But otherwise yes! Jealous much!

 

[tiny-tim]

Oh. That's going to be a long integration -.-

nooo … two short integrations …

you can split it up into two separate integrals, one over r and one over θ.

 

[ProPatto16]

But I thought to do that you had to have both integrating variables separate? The variable r is in both functions and cannot be factored out of either?

Or do you mean integrate with respect to r then respect to theta?

 

[tiny-tim]

But I thought to do that you had to have both integrating variables separate? The variable r is in both functions and cannot be factored out of either?

No, the integrand is a product (which can be separated), and the limits are independent …

what are the limits? ​

 

[ProPatto16]

Limits are 0 and a for one and the other one limits are 0 and the -ve root thing?

 

[tiny-tim]

nooo … the other one is for θ …

what are the limits for θ ? ​

 

[ProPatto16]

Intuitively, it would have to be between pi/2 and pi?

 

[tiny-tim]

Intuitively, it would have to be between pi/2 and pi?

Why "intuitively" ?

Yes, π/2 ≤ θ ≤ π …

the whole point of changing variables was to get the r and θ limits independent of each other!

 

[ProPatto16]

Intuition says for the quarter circle to be in upper left quadrant then that would have to be range of theta? Haha. Maybe not the right word. But made my point. And so then the integrand in respect to theta is evaluated with those bounds?

 

[tiny-tim]

yes!

sooo … ? ​

 

[ProPatto16]

So integrate with respect to theta then sub in bounds then integrate the solution with respect to r and sub in a?
Which would give a numerical answer multiplied by a?

 

[tiny-tim]

Which would give a numerical answer multiplied by a?

Do it and see!

(in fact, do it both ways …

integrate wrt θ first, then r, then try integrating wrt r first, then θ )​

 

[ProPatto16]

okay. i just typed up a huge long reply took me 45min and then i went to post and the page expired! so now I am soooo angry I am just going to go through it quick and hope you understand.

first thing. I am integrating wrt theta first, then r.

second thing, the integral of cos²theta.

integral of cos²x is 0.5x + 0.25sin(2x) so do i just replace the x with theta giving 0.5+0.25sin(2theta)? i will continue on having done that.

from there first i multiplied through the extra r that is in the end and then integrated wrt theta to get:

[(r³(0.5+0.25sin(2theta)))*(-r²cos(theta)] between pi and pi/2

evaluating that gives [(r³)(0.5)(r²), since some of the trig things are 0... which then becomes 0.5r⁵

now integrate that between a and 0

gives [r⁶/12] then just sub in the a giving a final answer of a⁶/12


the only two questions in that are can i multiply the r through at the start of do i have to do it during each integration?
and is that expression for cos²theta correct?

 

[ProPatto16]

"At the start or* do I have to.."
Wouldn't let me edit.

 

[tiny-tim]

Hi ProPatto16!

the only two questions in that are can i multiply the r through at the start of do i have to do it during each integration?
and is that expression for cos²theta correct?

Yes, cos²x = (1 + cos2x)/2, so the integral is x/2 + (sin2x)/4 + constant

I'm not sure what you mean about the r, but if you try integrating the other way round (wrt r first, then θ), you'll see that you get the same result.