MHB Double Integrals in Polar Coordinates

[harpazo]

Evaluate the double integral by converting to polar coordinates.

Let S S be the double integral symbol

S S xy dydx

Inner limits: 0 to sqrt{2x - x^2}

Outer limits: 0 to 2

The answer is 2/3.

I know that x = rcosϴ and y = rsinϴ.

S S rcosϴ*rsinϴ r drdϴ.

S S (r^3)cosϴ*sinϴ drdϴ.

I am stuck here.

I also need the correct limits of integration for this circle.

I just want the set up and explanation how to find the limits.

What must be done with the upper limit of integration for the inner integral?

 

[MarkFL]

The area over which we are integrating is the upper half of the circle centered at $(1,0)$, which is described in polar coordinates by:

$$r=2\cos(\theta)$$ where $$0\le\theta\le\frac{\pi}{2}$$ and $$0\le r\le2\cos(\theta)$$. And so the integral becomes:

$$I=\int_0^{\frac{\pi}{2}} \sin(\theta)\cos(\theta)\int_0^{2\cos(\theta)} r^3\,dr\,d\theta$$

Iterating the above integral will give the desired result. :D

 

[harpazo]

The area over which we are integrating is the upper half of the circle centered at $(1,0)$, which is described in polar coordinates by:

$$r=2\cos(\theta)$$ where $$0\le\theta\le\frac{\pi}{2}$$ and $$0\le r\le2\cos(\theta)$$. And so the integral becomes:

$$I=\int_0^{\frac{\pi}{2}} \sin(\theta)\cos(\theta)\int_0^{2\cos(\theta)} r^3\,dr\,d\theta$$

Iterating the above integral will give the desired result. :D

Where did the point (1,0) come from?

Where did r = 2cos(theta) come from?

 

[MarkFL]

Where did the point (1,0) come from?

Where did r = 2cos(theta) come from?

In the original integral, we have the inner upper limit of:

$$y=\sqrt{2x-x^2}$$

Squaring, we get:

$$y^2=2x-x^2$$

$$x^2-2x+y^2=0$$

$$x^2-2x+1+y^2=1$$

$$(x-1)^2+y^2=1^2$$

This is a circle centered at $(1,0)$, but we only want the upper half since in the original relation, $y$ is non-negative. To convert this to polar coordinates, let's return to:

$$y^2=2x-x^2$$

$$x^2+y^2=2x$$

$$r^2=2r\cos(\theta)$$

Divide through by $r$:

$$r=2\cos(\theta)$$

Notice that for $$\theta=\frac{\pi}{2}$$, we retain the solution $r=0$ that we lost when dividing by $r$. We also observe that for $$0\le\theta\le\frac{\pi}{2}$$, the upper half of the circle is traced out.

 

[harpazo]

In the original integral, we have the inner upper limit of:

$$y=\sqrt{2x-x^2}$$

Squaring, we get:

$$y^2=2x-x^2$$

$$x^2-2x+y^2=0$$

$$x^2-2x+1+y^2=1$$

$$(x-1)^2+y^2=1^2$$

This is a circle centered at $(1,0)$, but we only want the upper half since in the original relation, $y$ is non-negative. To convert this to polar coordinates, let's return to:

$$y^2=2x-x^2$$

$$x^2+y^2=2x$$

$$r^2=2r\cos(\theta)$$

Divide through by $r$:

$$r=2\cos(\theta)$$

Notice that for $$\theta=\frac{\pi}{2}$$, we retain the solution $r=0$ that we lost when dividing by $r$. We also observe that for $$0\le\theta\le\frac{\pi}{2}$$, the upper half of the circle is traced out.

I now understand what I did wrong. I noticed that you completed the square after squaring the upper limit.
I kept getting r^2 = 2x but it didn't dawn on me to replace x with rcos(theta). It makes sense to work out several more double integrals in polar coordinates before moving on to the next section dealing with center of mass and inertia.