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Math Help - Integration over a region

  1. #1
    Bar0n janvdl's Avatar
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    Integration over a region

    Hi guys,

    I'm a tad stuck on this problem.

    Let R be the region in \mathbb{R}^2 enclosed in the circle x^2+y^2 = 1 and below the line y = x + 1

    Calculate the integral \int \int _R x^2y^2 dA by order of integration dy \ dx
    First a sketch of the region:


    So I figured I could break it up into 2 regions, the triangular area under y=x+1 and one quarter of the circle which I can multiply by 3 to give me the total area of all 3 quarters.

    \int \int _R x^2y^2 dA = \int_{-1}^{0} \int_{0}^{x+1} x^2y^2 dy dx + 3 \int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} x^2y^2 dy dx

    My problem arises when trying to integrate the quarter of the circle.

    3 \int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} x^2y^2 dy dx = \int_{0}^{1} x^2 (\sqrt{1-x^2})^3 dx

    I've set x = sin(u) to transform the integral to:

    \int_{0}^{\frac{\pi}{2}} sin^2 (u) cos^3 (u) du = \int_{0}^{\frac{\pi}{2}} cos^3 (u) - cos^5 (u) du

    Which then by recursive use of the reduction formula gives me an answer of \frac{2}{15} which isn't correct since it should be \frac{3 \pi}{4}

    Where did I go wrong on this? Also, this question is only worth about 4 marks, and in my opinion my method is too much trouble for only 4 marks. What am I missing?

    All help appreciated.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by janvdl View Post
    Which then by recursive use of the reduction formula gives me an answer of \frac{2}{15} which isn't correct since it should be \frac{3 \pi}{4}

    Your result is correct. For example:


    \displaystyle\int_{0}^{\pi/2}\sin ^2x\cos^3x\;dx=\dfrac{1}{2}B(3/2,2)=\dfrac{1}{2}\dfrac{\Gamma(3/2)\Gamma(2)}{\Gamma (7/2)}=\ldots=\dfrac{2}{15}
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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by FernandoRevilla View Post
    Your result is correct. For example:


    \displaystyle\int_{0}^{\pi/2}\sin ^2x\cos^3x\;dx=\dfrac{1}{2}B(3/2,2)=\dfrac{1}{2}\dfrac{\Gamma(3/2)\Gamma(2)}{\Gamma (7/2)}=\ldots=\dfrac{2}{15}
    Thanks for the reply.

    I'm confused. The area of a circle is given by A = \pi r^2 If r=1 then A = \pi. But that's for the whole circle, so excluding the one quarter with the triangular area, it should be \frac{3}{4} \times A = \frac{3}{4} \pi

    So again, what am I missing?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by janvdl View Post
    Thanks for the reply.

    I'm confused. The area of a circle is given by A = \pi r^2 If r=1 then A = \pi. But that's for the whole circle, so excluding the one quarter with the triangular area, it should be \frac{3}{4} \times A = \frac{3}{4} \pi

    So again, what am I missing?

    Tha is right, but what has to do \iint_R x^2y^2\;dxdy with the area \iint_R \;dxdy ?.
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  5. #5
    Bar0n janvdl's Avatar
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    Quote Originally Posted by FernandoRevilla View Post
    Tha is right, but what has to do \iint_R x^2y^2\;dxdy with the area \iint_R \;dxdy ?.
    Of course! Sorry, I sat up until 2 AM with this problem, and I completely overlooked that.

    Thanks a lot!
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  6. #6
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by janvdl View Post
    Of course! Sorry, I sat up until 2 AM with this problem, and I completely overlooked that.

    Don't worry, he who is without sin among you, let him throw the first stone.
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