Results 1 to 10 of 10

Math Help - [SOLVED] A couple of double integrals

  1. #1
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318

    [SOLVED] A couple of double integrals

    EDIT: Solved all of them.

    I guess I don't know how to solve any of the following, because I keep getting the wrong answer all the time. It's mostly the boundaries that I'm having trouble with.

    1) \int{\int_D{x^3}}dxdy where D= \left\{(x,y)\in R^2; 1\leq x^2 + 9y^2 \leq 9; x \geq 3y\right\}


    2) \int{\int_D{x}}dxdy where D= \left\{(x,y)\in R^2; 1\leq x^2 + 2xy + 4y^2 \leq 1; x \geq 0; y \geq 0\right\}

    3) \int{\int_D{\frac{dxdy}{(1+x^2-y^2)^2}}}dxdy where D is the triangle with corners in (0,0), (1, -1) and (3,1).

    So here's what I tried to do:

    1)

    \left\{\begin{array}{l}u= x\\v = 3y\end{array}\right.

    |J(u,v)| = \frac{1}{3}

    1 \leq u^2 + v^2 \leq 9, u \geq v

    \left\{\begin{array}{c}u= rcos\varphi\\v = rsin\varphi\end{array}\right.

    |J(r, \varphi)| = r

    So the limits become 1 < r < 3 and \frac{-3\pi}{4}< \varphi < \frac{\pi}{4}
    Last edited by Spec; May 19th 2008 at 12:07 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Spec View Post
    I guess I don't know how to solve any of the following, because I keep getting the wrong answer all the time. It's mostly the boundaries that I'm having trouble with.

    1) \int{\int_D{x^3}}dxdy where D= \left\{(x,y)\in R^2; 1\leq x^2 + 9y^2 \leq 9; x \geq 3y\right\}

    [snip]

    So here's what I tried to do:

    1)

    \left\{\begin{array}{l}u= x\\v = \frac{y}{3}\end{array}\right.

    |J(x,y)| = \frac{1}{3}

    Boundaries?

    1 \leq u^2 + v^2 \leq 9

    \left\{\begin{array}{c}u= rcos\varphi\\v = rsin\varphi\end{array}\right.

    |J(r, \varphi)| = r

    Boundaries?

    \int{\int_D{r^3cos^3\varphi}}\cdot |J(u,v)| dxdy = \int{\int_D{\frac{r^4cos^3\varphi}{3}}}dxdy
    [snip]
    I have time for one:

    After your transformation to uv-coordinates (which was a good move by the way), you have  \frac{1}{3} \int \int_{D'} u^3 \, du \, dv where D' = \left\{(u ,v) \in R^2; 1 \leq u^2 + v^2 \leq 9; u \geq 9v \Rightarrow v \leq \frac{u}{9}\right\}.

    Drawing a diagram that shows D' will be a big help for you.

    Switch to polar coordinates and you have \frac{1}{3} \int_{r = 1}^{r = 3} \int_{\varphi = \alpha}^{\varphi = \beta} r^4 \cos^3\varphi \, d\varphi \, dr where \alpha = \tan^{-1} \left( \frac{1}{9} \right) - \pi and \beta = \tan^{-1} \left( \frac{1}{9} \right).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    I'm not getting the right answer by using that. It's supposed to be \frac{121\sqrt{2}}{9}, but I'm getting something way more complicated.

    I would appreciate some help with the other integrals as well.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Spec View Post
    I'm not getting the right answer by using that. It's supposed to be \frac{121\sqrt{2}}{9}, but I'm getting something way more complicated.

    I would appreciate some help with the other integrals as well.
    There's an error with the Jacobian. It's meant to be \left| \begin{array}{cc}<br />
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\<br />
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} \right| = \left| \begin{array}{cc}<br />
1 & 0 \\<br />
0 & 3 \end{array} \right| = 3.

    I can't see any other error. But I get \frac{5929 \sqrt{82}}{1681}.

    Maybe someone else can see what I can't (if there's anything to see).

    If I have time I'll take a look at the others. I would have thought that the second one is similar to the first in many ways ......
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by Spec View Post
    [snip]

    3) \int{\int_D{\frac{dxdy}{(1+x^2-y^2)^2}}}dxdy where D is the triangle with corners in (0,0), (1, -1) and (3,1).

    [snip]
    The triangular region D is bounded by the lines y = -x, ~ y = \frac{x}{3}\, and y = x - 2.

    Note that 1 + x^2 - y^2 = 1 + (x - y)(x + y). This suggests making the transformation

    u = x - y .... (1)

    v = x + y .... (2)

    (1) + (2): x = \frac{u+v}{2}.

    (2) - (1): y = \frac{-u+v}{2}.

    The Jacobian of the transformation is therefore J(u, v) = \left| \begin{array}{cc}<br />
\frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\<br />
 & \\<br />
\frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array} \right| = \left| \begin{array}{cc}<br />
\frac{1}{2} & \frac{1}{2} \\<br />
 & \\<br />
-\frac{1}{2} & \frac{1}{2} \end{array} \right| = \frac{1}{2}.

    The triangular region D transforms to a simple triangular region D' bounded by the following lines:

    y = - x \Rightarrow x + y = 0 \Rightarrow {\color{red}v = 0}.

    y = x - 2 \Rightarrow x - y = 2 \Rightarrow {\color{red}u = 2}.

    y = \frac{x}{3} \Rightarrow \frac{-u + v}{2} = \frac{u + v}{6} \Rightarrow {\color{red}v = 2u}.

    So the integral becomes \frac{1}{2} \int_{u=0}^{u=2} \int_{v = 0}^{v = 2u} \frac{1}{(1 + uv)^2} \, dv \, du and it should be blue sky from here.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Quote Originally Posted by mr fantastic View Post
    \alpha = \tan^{-1} \left( \frac{1}{9} \right) - \pi
    \beta = \tan^{-1} \left( \frac{1}{9} \right)
    Why are they \tan^{-1} \left( \frac{1}{9} \right) ?


    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by wingless View Post
    Why are they \tan^{-1} \left( \frac{1}{9} \right) ?


    In transforming from xy-coordinates to uv-coordinates, the line x = 3y becomes u = 3(3v) = 9v => v = u/9, which has a gradient of 1/9 ......

    The transformation is done to facilitate a switch to polar coordinates.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Quote Originally Posted by Spec View Post
    I'm not getting the right answer by using that. It's supposed to be \frac{121\sqrt{2}}{9}, but I'm getting something way more complicated.

    I would appreciate some help with the other integrals as well.

    I got \frac{121\sqrt{2}}{9}. Here's my approach:

    \int{\int_D{x^3}}dxdy where D= \left\{(x,y)\in R^2; 1\leq x^2 + 9y^2 \leq 9; x \geq 3y\right\}

    Let x = r \cos \theta and y = \frac{1}{3}r \sin \theta. Then
    D'=1\leq r^2 \leq 9;~\cos\theta \geq \sin\theta.

    \int\int_D x^3~dx~dy = \int\int_{D'}r^3\cos^3 \theta \frac{r}{3}~dr~d\theta.

    From 1\leq r^2 \leq 9,~1\leq r \leq 3.
    From \cos\theta \geq \sin\theta,~-\frac{3\pi}{4}\leq \theta \leq \frac{\pi}{4}.

    \int^{\frac{\pi}{4}}_{-\frac{3\pi}{4}}\int_1^ 3\frac{1}{3}r^4\cos^3 \theta ~dr~d\theta = \frac{121 \sqrt{2}}{9}.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by wingless View Post
    I got \frac{121\sqrt{2}}{9}. Here's my approach:

    \int{\int_D{x^3}}dxdy where D= \left\{(x,y)\in R^2; 1\leq x^2 + 9y^2 \leq 9; x \geq 3y\right\}

    Let x = r \cos \theta and y = \frac{1}{3}r \sin \theta. Then
    D'=1\leq r^2 \leq 9;~\cos\theta \geq \sin\theta.

    \int\int_D x^3~dx~dy = \int\int_{D'}r^3\cos^3 \theta \frac{r}{3}~dr~d\theta.

    From 1\leq r^2 \leq 9,~1\leq r \leq 3.
    From \cos\theta \geq \sin\theta,~-\frac{3\pi}{4}\leq \theta \leq \frac{\pi}{4}.

    \int^{\frac{\pi}{4}}_{-\frac{3\pi}{4}}\int_1^ 3\frac{1}{3}r^4\cos^3 \theta ~dr~d\theta = \frac{121 \sqrt{2}}{9}.
    The one thing I didn't check - Spec's transformation from xy-coordinates to uv-coordinates. I used the original transformation (see the quote in post #2) without checking it

    Looking for my mistake, I see that it should have been v = 3y, NOT v = y/3. Then I notice Spec has edited the question to reflect this - s/he obviously realised the same thing (*ahem* it's good form to include the reason for an edit, by the way .....)

    My mistake for not checking right from the start I knew it would be something simple.

    Moral of the lesson - don't assume that even the simplest things are done correctly.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    Quote Originally Posted by mr fantastic View Post
    Moral of the lesson - don't assume that even the simplest things are done correctly.
    That's so true
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. a couple of trigonometric integrals
    Posted in the Calculus Forum
    Replies: 7
    Last Post: December 27th 2010, 08:43 AM
  2. Need help with couple of integrals!
    Posted in the Calculus Forum
    Replies: 9
    Last Post: June 1st 2010, 11:32 AM
  3. Replies: 4
    Last Post: April 11th 2010, 09:39 PM
  4. a couple integrals
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 18th 2008, 01:20 PM
  5. A couple of indefinite integrals
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 18th 2006, 01:29 AM

Search Tags


/mathhelpforum @mathhelpforum