# Tricky Double Integral with polar coordinates

#### craig

Evaluate the integral:

$$\displaystyle \int \int_D \frac{x^2}{x^2 + y^2} dx\;dy$$

where $$\displaystyle D$$ is the region $$\displaystyle \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$$

Tried drawing the region $$\displaystyle D$$ and it looks like I have to use polar coordinates.

Using polar coordinates, $$\displaystyle \frac{x^2}{x^2 + y^2} = \frac{r^2 \cos^2{\theta}}{r^2} = \cos^2{\theta}$$.

I'm not sure how to find the limits if integration though, or what to do with the $$\displaystyle dx\;dy$$.

Thanks in advance, any pointers would be greatly appreciated.

#### Mush

Evaluate the integral:

$$\displaystyle \int \int_D \frac{x^2}{x^2 + y^2} dx\;dy$$

where $$\displaystyle D$$ is the region $$\displaystyle D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$$

Tried drawing the region $$\displaystyle D$$ and it looks like I have to use polar coordinates.

Using polar coordinates, $$\displaystyle \frac{x^2}{x^2 + y^2} = \frac{r^2 \cos^2{\theta}}{r^2} = \cos^2{\theta}$$.

I'm not sure how to find the limits if integration though, or what to do with the $$\displaystyle dx\;dy$$.

Thanks in advance, any pointers would be greatly appreciated.
In general to convert, you do the following:

$$\displaystyle \displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta) \frac{\partial (x,y)}{\partial (r, \theta)} \, dr \, d\theta$$

Where $$\displaystyle \frac{\partial (x,y)}{\partial (r, \theta)}$$ is called the jacobian determinant and it's given by:

$$\displaystyle \frac{\partial (x,y)}{\partial (r, \theta)} = \begin{vmatrix} \cos \theta & - r \sin \theta \\ \sin \theta & r \cos \theta \end{vmatrix} = r$$

Hence:

$$\displaystyle \displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta) r \, dr \, d\theta$$

Now, looking at the region in rectilinear coordinates can give you some hints as to the limits in polar coordinates.

$$\displaystyle D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$$

Now, the 2nd constraint on the region says that $$\displaystyle x^2 + y^2 \leq 3$$. You should be able to see that this equation describes a circle whose radius is less than $$\displaystyle \sqrt{3}$$, in other words, the radius is limited by $$\displaystyle 0 \leq r \leq \sqrt{3}$$.

Now, the 1st constraint on the region says that $$\displaystyle y \geq x$$. Now, we know that the line at which y= x is a diagonal line going through the origin. And y is greater than x in any point that lies ABOVE that line. In other words, the region in question is made up of points above or on the line y = x. This means that your theta is limited by $$\displaystyle \frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4}$$.

You can see all of this visually by drawing it out. Go on, do this:

1) Draw the x-y coordinates.
2) Draw a circle centred at the origin with a radius of $$\displaystyle \sqrt{3}$$.
3) Draw the line y = x.
4) Now shade the region that lies above the line y = x, but does not go outside the circle.

You should see that this region is bounded by:

$$\displaystyle D = \bigg\{ (r,\theta)| 0 \leq r \leq \sqrt{3} \; \textrm{and} \; \frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4} \bigg\}$$

Hence, in this particular situation:

$$\displaystyle \int \int_D \frac{x^2}{x^2 + y^2} dx\;dy = \int_0^{\frac{5\pi}{4}} \int_0^{\sqrt{3}} r \, \cos^2(\theta) \, dr \, d\theta = \int_0^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$$

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• craig

#### running-gag

• craig

#### AllanCuz

Evaluate the integral:

$$\displaystyle \int \int_D \frac{x^2}{x^2 + y^2} dx\;dy$$

where $$\displaystyle D$$ is the region $$\displaystyle \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$$

Tried drawing the region $$\displaystyle D$$ and it looks like I have to use polar coordinates.

Using polar coordinates, $$\displaystyle \frac{x^2}{x^2 + y^2} = \frac{r^2 \cos^2{\theta}}{r^2} = \cos^2{\theta}$$.

I'm not sure how to find the limits if integration though, or what to do with the $$\displaystyle dx\;dy$$.

Thanks in advance, any pointers would be greatly appreciated.
I go through a very similar problem here: http://www.mathhelpforum.com/math-help/519249-post2.html

If you draw the xy domain you can see we want the area to the left of the line but inside the circle,

If we convert this to polar co-ordinates this means

$$\displaystyle \frac{ \pi }{4} \le \theta \le \frac{ 5 \pi }{4}$$ where the $$\displaystyle \frac{ \pi }{4}$$ comes from the fact that the line $$\displaystyle y=x$$ acts at $$\displaystyle \frac{ \pi }{4}$$ above the x-axis. And $$\displaystyle \frac{ 5 \pi }{4}$$ comes from the angle between the line in the first quadrent to the line in the 3rd quadrent.

For r, we are going from the origin to the radius of the circle. So the interval of r must be

$$\displaystyle 0 \le r \le \sqrt{3}$$

We can now compute this integral,

$$\displaystyle \int \int_D \frac{x^2}{x^2 + y^2} dxdy$$

$$\displaystyle = \iint_Q \frac{ r^2 cos^2 \theta }{ r^2 } rdr d \theta$$

$$\displaystyle = \iint_Q rcos^2 \theta dr d \theta$$

$$\displaystyle = \int_0^{ \sqrt{3} } rdr \int_{ \frac{ \pi }{4} }^{ \frac{ 5 \pi }{4} } cos^2 \theta d \theta$$

Which can be easily computed using double angle formulas Last edited:
• craig

#### craig

Wow leave the forum for a few hours and there's 3 great replies waiting for me!!

In general to convert, you do the following:

$$\displaystyle \displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta) \frac{\partial (x,y)}{\partial (r, \theta)} \, dr \, d\theta$$

Where $$\displaystyle \frac{\partial (x,y)}{\partial (r, \theta)}$$ is called the jacobian determinant and it's given by:

$$\displaystyle \frac{\partial (x,y)}{\partial (r, \theta)} = \begin{vmatrix} \cos \theta & - r \sin \theta \\ \sin \theta & r \cos \theta \end{vmatrix} = r$$

Hence:

$$\displaystyle \displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta) r \, dr \, d\theta$$
Ahh yeh I've done Jacobian determinants before, it didn't click that it applied here.

Now, looking at the region in rectilinear coordinates can give you some hints as to the limits in polar coordinates.

$$\displaystyle D = \{ (x,y)| y \geq x \; \textrm{and} \; x^2 + y^2 \leq 3 \}$$

Now, the 2nd constraint on the region says that $$\displaystyle x^2 + y^2 \leq 3$$. You should be able to see that this equation describes a circle whose radius is less than $$\displaystyle \sqrt{3}$$, in other words, the radius is limited by $$\displaystyle 0 \leq r \leq \sqrt{3}$$.

Now, the 1st constraint on the region says that $$\displaystyle y \geq x$$. Now, we know that the line at which y= x is a diagonal line going through the origin. And y is greater than x in any point that lies ABOVE that line. In other words, the region in question is made up of points above or on the line y = x. This means that your theta is limited by $$\displaystyle \frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4}$$.

You can see all of this visually by drawing it out. Go on, do this:

1) Draw the x-y coordinates.
2) Draw a circle centred at the origin with a radius of $$\displaystyle \sqrt{3}$$.
3) Draw the line y = x.
4) Now shade the region that lies above the line y = x, but does not go outside the circle.

You should see that this region is bounded by:

$$\displaystyle D = \bigg\{ (r,\theta)| 0 \leq r \leq \sqrt{3} \; \textrm{and} \; \frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4} \bigg\}$$
I'd drawn the circle but for some reason only drew $$\displaystyle y = x$$ for positive values of $$\displaystyle x$$ and $$\displaystyle y$$, that would be where I was going wrong I guess Hence, in this particular situation:

$$\displaystyle \int \int_D \frac{x^2}{x^2 + y^2} dx\;dy = \int_0^{\frac{5\pi}{4}} \int_0^{\sqrt{3}} r \, \cos^2(\theta) \, dr \, d\theta = \int_0^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$$
** Think you mean $$\displaystyle \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$$

Anyway on with the integral:

$$\displaystyle \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$$

$$\displaystyle \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg[ \frac{r^2}{2} \bigg]_0^{\sqrt{3}} \, d\theta$$ = $$\displaystyle \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \frac{1}{2}(1 + \cos{2\theta}) \, d\theta$$ =

$$\displaystyle \frac{3}{4}\bigg[\theta + \frac{1}{2}\sin{2\theta}\bigg]_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$$ =

$$\displaystyle \frac{3}{4}\bigg(\frac{5\pi}{4} + \frac{1}{2}\sin{\frac{5\pi}{2}} -\frac{pi}{4} - \frac{1}{2}\sin{\frac{\pi}{2}} \bigg)$$

$$\displaystyle \frac{3}{4}(\pi + 0)$$ = $$\displaystyle \frac{3\pi}{4}$$

Thanks a lot for the help!

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#### craig

Hi

The area of integration is the part of the disk which is above the red line And $$\displaystyle dx dy$$ becomes $$\displaystyle r dr d\theta$$
Thanks a lot for the image, I'd forgot to draw the whole of the line $$\displaystyle y=x$$!

I go through a very similar problem here: http://www.mathhelpforum.com/math-help/519249-post2.html

If you draw the xy domain you can see we want the area under the line but inside the circle,

If we convert this to polar co-ordinates this means

$$\displaystyle 0 \le \theta \le \frac{ \pi }{4}$$ where the $$\displaystyle \frac{ \pi }{4}$$ comes from the fact that the line $$\displaystyle y=x$$ acts at $$\displaystyle \frac{ \pi }{4}$$ above the x-axis.

For r, we are going from the origin to the radius of the circle. So the interval of r must be

$$\displaystyle 0 \le r \le \sqrt{3}$$

We can now compute this integral,

$$\displaystyle \int \int_D \frac{x^2}{x^2 + y^2} dxdy$$

$$\displaystyle = \iint_Q \frac{ r^2 cos^2 \theta }{ r^2 } rdr d \theta$$

$$\displaystyle = \iint_Q rcos^2 \theta dr d \theta$$

$$\displaystyle = \int_0^{ \sqrt{3} } rdr \int_0^{ \frac{ \pi }{4} } cos^2 \theta d \theta$$

Which can be easily computed using double angle formulas Thanks a lot for the reply, really helped.

#### Mush

Wow leave the forum for a few hours and there's 3 great replies waiting for me!!

Ahh yeh I've done Jacobian determinants before, it didn't click that it applied here.

I'd drawn the circle but for some reason only drew $$\displaystyle y = x$$ for positive values of $$\displaystyle x$$ and $$\displaystyle y$$, that would be where I was going wrong I guess ** Think you mean $$\displaystyle \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$$

Anyway on with the integral:

$$\displaystyle \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg( \int_0^{\sqrt{3}} r \, \, dr\bigg) \, d\theta$$

$$\displaystyle \int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \cos^2(\theta) \bigg[ \frac{r^2}{2} \bigg]_0^{\sqrt{3}} \, d\theta$$ = $$\displaystyle \frac{3}{2}\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} \frac{1}{2}(1 + \cos{2\theta}) \, d\theta$$ =

$$\displaystyle \frac{3}{4}\bigg[1 + \cos{2\theta}\bigg]_{\frac{\pi}{4}}^{\frac{5\pi}{4}}$$ = $$\displaystyle \frac{3}{4}(\pi)$$ = $$\displaystyle \frac{3\pi}{4}$$

Thanks a lot for the help!
I don't think you're performed that integration properly, have you?

The integral of $$\displaystyle 1 + \cos(2\theta)$$ is $$\displaystyle \theta + \frac{1}{2}\sin(2\theta)$$.

#### AllanCuz

Thanks a lot for the image, I'd forgot to draw the whole of the line $$\displaystyle y=x$$!

Thanks a lot for the reply, really helped.
I messed up my bounds for theta. I thought the question said $$\displaystyle y = x$$ but in fact, we want $$\displaystyle y \ge x$$ which is the area inside the circle and to the left of the line. My post has been edited to explain the angles. Sorry about that #### craig

I don't think you're performed that integration properly, have you?
Yes & no. I did the integration properly, just forgot to update the latex code...oops.

Edited my above working now, sorry about that.

#### craig

I messed up my bounds for theta. I thought the question said $$\displaystyle y = x$$ but in fact, we want $$\displaystyle y \ge x$$ which is the area inside the circle and to the left of the line. My post has been edited to explain the angles. Sorry about that Haha thought that was what you meant, thanks again for the reply.