Converting into polar coordinates for integration

[VinnyCee]

Another problem that I cannot figure out. Convert the follorwing into polar coordinates:

$$\int_{-1}^{1} \int_{-\sqrt{1 - y^2}}^{\sqrt{1 - y^2}} ln\left(x^2 + y^2 + 1\right) dx\;dy$$

I did this so far:

$$ln\left(x^2 + y^2 + 1\right) = ln\left(r^2 + 1\right)$$

$$\sqrt{1 - y^2} = \sqrt{1 - r^2 \sin^2 \theta}$$

Now what do I do?

Is this possibly right?

$$\int_{0}^{2\pi} \int_{-\sqrt{1 - r^2 \sin^2 \theta}}^{\sqrt{1 - r^2 \sin^2 \theta}} ln\left(r^2 + 1\right) dr\;d\theta$$

 

[whozum]

Trig identities, what does [itex] 1-sin^2(\theta) [/tex] equal?

 

[whozum]

Better yet,

x^2+y^2 = r^2
with r = 1,
x^2+y^2 = 1
x^2 = 1-y^2

x = sqrt(1-y^2)

 

[SpaceTiger]

If what whozum said isn't clear to you, try drawing a graph of the region you're integrating over. Also, remember that the area element for a polar integral is

$$dA=rdrd\theta$$

 

[VinnyCee]

Is this it?

Including the $$dA = r\;dr\;d\theta$$, is this integral the right conversion?

$$\int_{0}^{2\pi} \int_{-\sqrt{1 - r^2 \sin^2 \theta}}^{\sqrt{1 - r^2 \sin^2 \theta}} r\;ln\left(r^2 + 1\right) dr\;d\theta$$

 

[whozum]

The limits of r simplify, other than that your right. Look at our hints above.

 

[VinnyCee]

Cool , but how does the $$1-sin^2(\theta) = \cos^2\left(\theta\right)$$ help when I have these limits of integration for $r$?

$$\sqrt{1 - r^2 \sin^2 \theta}}$$

It does not factor right?

Attached is a graph.

 

[SpaceTiger]

$$\int_{0}^{2\pi} \int_{-\sqrt{1 - r^2 \sin^2 \theta}}^{\sqrt{1 - r^2 \sin^2 \theta}} r\;ln\left(r^2 + 1\right) dr\;d\theta$$

Does it make sense to have "r" in the limit of an integral over r?

 

[whozum]

It does factor, that's my point. Read what I said above clearly, sketch the domain. It will be clear what your limits for R are.

 

[VinnyCee]

Let's try this?

$$x = \sqrt{1 - y^2}$$
$$r\;\cos\theta = \sqrt{1 - r^2\;\sin^2\theta}$$

Then solve for r?

 

[VinnyCee]

Like this?

$$r^2\;\cos^2\theta = 1 - r^2\;\sin^2\theta$$

$$r^2\;\left(\cos^2\theta + \sin^2\theta\right) = 1$$

$$r^2 = 1$$

$r$ limits are then from $0$ to $1$?

The right integral?

$$\int_{0}^{2\pi} \int_{0}^{1} r\;ln\left(r^2 + 1\right) dr\;d\theta$$

 

[whozum]

Correct, though it might be -1 to 1, spacetiger will give you the final ok.

 

[HallsofIvy]

The point of all the comments was : DRAW A PICTURE!
$$y= \sqrt{1- x^2}$$ and $$y=-\sqrt{1- x^2}$$ both are the same as $$x^2= 1- y^2$$ or $$x^2+ y^2= 1$$, the unit circle. Since x ranges from -1 to 1, the region of integration is simply the unit circle. To cover that, $$\theta$$ ranges from 0 to $$2\pi$$ and r from 0 to 1.