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Math Help - Double Integrals

  1. #1
    Super Member Aryth's Avatar
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    Double Integrals

    There are two of them that I can't figure out:

    1. \int^2_0\int^{\frac{\pi}{2}}_0 x\sin{y} \ dydx

    On this one I got 0, but it's supposed to be 2

    And

    2. \int^2_0\int^1_0 (2x + y)^8 \ dxdy

    On this one I got \frac{262144}{45} but its supposed to be \frac{261632}{45}. My answer is 2^9 off.

    Any help is appreciated.
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  2. #2
    Super Member wingless's Avatar
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    \int^2_0\int^{\frac{\pi}{2}}_0 x\sin{y} \ dydx

    \int^2_0 x~dx \int^{\frac{\pi}{2}}_0 \sin y~dy

    I think you can do it now.
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  3. #3
    Eater of Worlds
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    \int_{0}^{2}\int_{0}^{\frac{\pi}{2}}xsin(y)dydx

    The first integration yields:

    \int xsin(y)dy=-xcos(y)

    Now, using the limits:

    -xcos(\frac{\pi}{2})+xcos(0)=x

    Now for that part:

    \int_{0}^{2}x dx

    Of course, this is easy enough:

    \frac{1}{2}(2)^{2}-\frac{1}{2}(0)^{2}=\boxed{2}
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  4. #4
    Super Member Aryth's Avatar
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    Thanks. Didn't consider the cos(0).

    Thanks wingless... I forgot about splitting the functions.

    Anyone for the second problem?
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  5. #5
    Math Engineering Student
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    Make the substitution z=2x+y for the inner integral, the rest is just straightforward computation.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Aryth View Post
    Thanks. Didn't consider the cos(0).

    Thanks wingless... I forgot about splitting the functions.

    Anyone for the second problem?
    the second one we can do by substitution.

    first letting u = 2x + y, we get: \int_0^2 \frac 1{18} (2x + y)^9 \bigg|_0^1 ~dy

    continue in the same manner. was that what you tried?
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  7. #7
    Super Member Aryth's Avatar
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    Ah, I did it the same way, but I see what I did wrong... It was in the first iteration:

    \frac{1}{18}(2x + y)^9\bigg|_0^1 = \frac{1}{18}\left[(2 + y)^9 - {\color{red}{y^9}}\right]

    That would completely explain why I was 2^9 over. I completely forgot the lower limit.
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Aryth View Post
    Ah, I did it the same way, but I see what I did wrong... It was in the first iteration:

    \frac{1}{18}(2x + y)^9\bigg|_0^1 = \frac{1}{18}\left[(2 + y)^9 - {\color{red}{y^9}}\right]

    That would completely explain why I was 2^9 over.
    ah, ok, good catching your mistake
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  9. #9
    Super Member Aryth's Avatar
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    Thanks... Bad news is, I made the same mistake on both problems. I gotta watch the lower limits. :P
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Aryth View Post
    Thanks... Bad news is, I made the same mistake on both problems. I gotta watch the lower limits. :P
    haha, indeed. now you know you weakness, you can work at squashing it. this should not hurt you in an exam now
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