Results 1 to 7 of 7

Math Help - double integrals in polar coordinates

  1. #1
    Newbie
    Joined
    Dec 2010
    From
    Tennessee
    Posts
    4

    double integrals in polar coordinates

    I have to use a double integral in polar coordinates to find the area of the region inside the circle x^2 + y^2 = 4 and to the right of the line x =1.

    I have it all graphed and I know how to do double integrals but I am confused as to what I am integrating, it is probably really simple but I was sick the day we went over this section in class so I have no idea what I am doing.

    Please help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,560
    Thanks
    1425
    Start by writing your double integral in cartesians first.

    First, where do the two graphs intersect? When \displaystyle x = 1.

    So \displaystyle 1^2 + y^2 = 4

    \displaystyle 1 + y^2 = 4

    \displaystyle y^2 = 3

    \displaystyle y = \pm \sqrt{3}.


    So if you take horiztonal strips, your \displaystyle x values range from \displaystyle 1 to the graph \displaystyle x = \sqrt{4 - y^2}, and your \displaystyle y values range from \displaystyle -\sqrt{3} to \displaystyle \sqrt{3}.

    So what is your double integral in cartesians?
    Last edited by Prove It; December 9th 2010 at 02:41 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2010
    From
    Tennessee
    Posts
    4

    double integrals in cartesian coordinates

    Here's a picture of what I have so far, I hope I am going in the right direction:
    double integrals in polar coordinates-img_0053.jpg

    would i replace x^2 + y^2 with r? if so what would my boundaries be? I'm confused by this....
    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 danyellogram View Post
    I have to use a double integral in polar coordinates to find the area of the region inside the circle x^2 + y^2 = 4 and to the right of the line x =1.

    I have it all graphed and I know how to do double integrals but I am confused as to what I am integrating, it is probably really simple but I was sick the day we went over this section in class so I have no idea what I am doing.

    Please help
    For future problems you can reference my tutorial on double integration at the top of this forum.

    I'll offer a different approach to Prove It (actually it's the same approach just worded differently) and outline my approach via steps so that you can follow the same process for other questions. Though I wouldn't write this in cartesian coordinates before moving to polar, I would go straight to polar.

    Step 1: Draw the Domain
    In this case we have a circle of radius 2 and a vertical line at x = 1. If we want the area to the right of this then we are looking at two quarter circles. For simplicity we will compute the area of 1 of these quarters and multiply by 2.

    Step 2: Set up the integral in general terms
     \iint_D f(x) dA

    Note that before we can approach this problem we need to find a set of bounds. Well, these bounds depend on which way we want to integrate our function.

    This actually has some complexity to it because we are computing this integral in polar co-ordinates.

    Step 3: Find Bounds
    Let's look at the area of a regular circle of radius 2.

     A = \int_0^{2 \pi } d \theta \int_0^2 rdr = \pi (2^2)

    But in our case we have a nice line going right up the middle and our radius is no longer constant. If you draw a line from the origin to the radius of the circle you'll see that it crosses the line x=1. Well, we dont want anything before that line so if we make a triangle we see that

     cos \theta = \frac{ 1 }{R} \to R= \frac{1}{cos \theta}

    Well this is good, so we now have our bounds for R

     \frac{1}{cos \theta } \le R \le 2

    We now need to evaluate theta. Well clearly we start at 0 (defined from the x-axis) but where do we end? Well we end where the line x=1 intersects the circle. This happens when x=1 so when we sub x=1 into our circle, we get the y bound of  y = \sqrt{3} . Now to get the theta angle we have a triangle with a height of square root 3 and a width of 1.

     Tan \theta = \frac{ \sqrt{3} }{1} \to \theta = 60

    So theta runs from 0 to 60.

    Step 4: Put it all together
     A = 2 \int_0^{ \frac{ \pi }{3} } d \theta \int_{ \frac{1}{cos \theta} }^2 r dr
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2010
    From
    Tennessee
    Posts
    4
    thank you both for your help! <3
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,560
    Thanks
    1425
    Quote Originally Posted by danyellogram View Post
    Here's a picture of what I have so far, I hope I am going in the right direction:
    Click image for larger version. 

Name:	IMG_0053.jpg 
Views:	34 
Size:	357.4 KB 
ID:	20042

    would i replace x^2 + y^2 with r? if so what would my boundaries be? I'm confused by this....
    I also realise I have made a mistake which you have taken through to your working - the \displaystyle x values range from \displaystyle +1 to \displaystyle \sqrt{4 - y^2}.


    Also, if you are taking horizontal strips, then you are integrating with respect to \displaystyle x first. So that means the double integral should be

    \displaystyle \int_{-\sqrt{3}}^{\sqrt{3}}{\int_{1}^{\sqrt{4-y^2}}{1\,dx}\,dy}. Now follow Allan's steps to convert to polars.


    Epilogue: You don't even need to convert to polars:

    \displaystyle \int_{-\sqrt{3}}^{\sqrt{3}}{\int_{1}^{\sqrt{4-y^2}}{\,dx}\,dy} = \int_{-\sqrt{3}}^{\sqrt{3}}{\left[x\right]_1^{\sqrt{4-y^2}}\,dy}

    \displaystyle = \int_{-\sqrt{3}}^{\sqrt{3}}{\sqrt{4 - y^2} - 1\,dy}

    \displaystyle = \int_{-\sqrt{3}}^{\sqrt{3}}{\sqrt{4 - y^2}\,dy} - \int_{-\sqrt{3}}^{\sqrt{3}}{1\,dy}

    Now you can make a trigonometric substitution to solve.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Dec 2010
    From
    Tennessee
    Posts
    4
    Yes when I was working it out (before allan's post) I found that slight mistake

    again, thank you for all of your help.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. double integrals in polar coordinates
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 21st 2010, 10:57 PM
  2. Replies: 3
    Last Post: November 21st 2009, 12:57 PM
  3. Polar coordinates and double integrals
    Posted in the Calculus Forum
    Replies: 5
    Last Post: June 5th 2009, 04:06 AM
  4. Double Integrals w/ Polar Coordinates
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 10th 2009, 01:30 AM
  5. Double integrals into polar coordinates
    Posted in the Calculus Forum
    Replies: 5
    Last Post: May 2nd 2008, 07:26 AM

Search Tags


/mathhelpforum @mathhelpforum