Results 1 to 11 of 11

Math Help - Tricky Double Integral with polar coordinates

  1. #1
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1

    Tricky Double Integral with polar coordinates

    Evaluate the integral:

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

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

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

    Using polar coordinates, \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 dx\;dy.

    Thanks in advance, any pointers would be greatly appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by craig View Post
    Evaluate the integral:

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

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

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

    Using polar coordinates, \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 dx\;dy.

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

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

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

    \frac{\partial (x,y)}{\partial (r, \theta)} =<br />
\begin{vmatrix}<br />
\cos \theta & - r \sin \theta \\<br />
\sin \theta & r \cos \theta<br />
\end{vmatrix} = r

    Hence:


     \displaystyle \int \int f(x,y) \, dx \, dy = \int \int f(r,\theta) <br />
r \, dr \, d\theta

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

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

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

    Now, the 1st constraint on the region says that  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  \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 \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:

     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:

    \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
    Last edited by Mush; May 30th 2010 at 08:14 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Hi

    The area of integration is the part of the disk which is above the red line



    And dx dy becomes r dr d\theta
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member AllanCuz's Avatar
    Joined
    Apr 2010
    From
    Canada
    Posts
    384
    Thanks
    4
    Quote Originally Posted by craig View Post
    Evaluate the integral:

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

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

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

    Using polar coordinates,  \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 dx\;dy.

    Thanks in advance, any pointers would be greatly appreciated.
    I go through a very similar problem here: http://www.mathhelpforum.com/math-he...249-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

      \frac{ \pi }{4} \le \theta \le \frac{ 5 \pi }{4} where the  \frac{ \pi }{4} comes from the fact that the line  y=x acts at  \frac{ \pi }{4} above the x-axis. And  \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

     0 \le r \le \sqrt{3}

    We can now compute this integral,

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

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

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

     = \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 by AllanCuz; May 30th 2010 at 01:37 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Wow leave the forum for a few hours and there's 3 great replies waiting for me!!

    Quote Originally Posted by Mush View Post
    In general to convert, you do the following:

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

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

    \frac{\partial (x,y)}{\partial (r, \theta)} =<br />
\begin{vmatrix}<br />
\cos \theta & - r \sin \theta \\<br />
\sin \theta & r \cos \theta<br />
\end{vmatrix} = r

    Hence:


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

    Quote Originally Posted by Mush View Post
    Now, looking at the region in rectilinear coordinates can give you some hints as to the limits in polar coordinates.

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

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

    Now, the 1st constraint on the region says that  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  \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 \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:

     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 y = x for positive values of x and y, that would be where I was going wrong I guess

    Quote Originally Posted by Mush View Post
    Hence, in this particular situation:

    \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 \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:


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


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

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

    \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)

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

    Thanks a lot for the help!
    Last edited by craig; May 30th 2010 at 01:39 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Quote Originally Posted by running-gag View Post
    Hi

    The area of integration is the part of the disk which is above the red line



    And dx dy becomes r dr d\theta
    Thanks a lot for the image, I'd forgot to draw the whole of the line y=x!


    Quote Originally Posted by AllanCuz View Post
    I go through a very similar problem here: http://www.mathhelpforum.com/math-he...249-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

     0 \le \theta \le \frac{ \pi }{4} where the  \frac{ \pi }{4} comes from the fact that the line  y=x acts at  \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

     0 \le r \le \sqrt{3}

    We can now compute this integral,

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

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

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

     = \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.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by craig View Post
    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 y = x for positive values of x and y, that would be where I was going wrong I guess



    ** Think you mean \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:


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


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

    \frac{3}{4}\bigg[1 + \cos{2\theta}\bigg]_{\frac{\pi}{4}}^{\frac{5\pi}{4}} = \frac{3}{4}(\pi) = \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  1 + \cos(2\theta) is  \theta + \frac{1}{2}\sin(2\theta) .
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member AllanCuz's Avatar
    Joined
    Apr 2010
    From
    Canada
    Posts
    384
    Thanks
    4
    Quote Originally Posted by craig View Post
    Thanks a lot for the image, I'd forgot to draw the whole of the line y=x!




    Thanks a lot for the reply, really helped.
    I messed up my bounds for theta. I thought the question said  y = x but in fact, we want  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 :P
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Quote Originally Posted by Mush View Post
    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.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    Quote Originally Posted by AllanCuz View Post
    I messed up my bounds for theta. I thought the question said  y = x but in fact, we want  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 :P
    Haha thought that was what you meant, thanks again for the reply.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by craig View Post
    Yes & no. I did the integration properly, just forgot to update the latex code...oops.

    Edited my above working now, sorry about that.
    Indeed. Good show.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Double Integral In Polar Coordinates
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 13th 2010, 01:48 AM
  2. Double integral of polar coordinates
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 22nd 2010, 09:16 AM
  3. double integral using polar coordinates
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 18th 2009, 01:14 AM
  4. Double integral and polar coordinates
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 27th 2008, 07:07 PM
  5. Double Integral in Polar Coordinates
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 18th 2008, 04:13 PM

Search Tags


/mathhelpforum @mathhelpforum