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Math Help - Evaluate

  1. #1
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    Evaluate

    semicircle bounded by x-axis and the curve y=sqrt(1-x^2)
    double int[e^((x^2)+(y^2))]dydx
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  2. #2
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    Quote Originally Posted by chevy900ss View Post
    semicircle bounded by x-axis and the curve y=sqrt(1-x^2)
    double int[e^((x^2)+(y^2))]dydx

    Polar coordinates: x=r\cos\theta\,,\,y=r\sin\theta\,,\,\,0\leq \theta\leq \pi\,,\,\,0\leq r\leq 1, so:

    \int\limits_0^{\pi}d\theta \int\limits_0^1re^{r^2}dr=...

    Tonio
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  3. #3
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    is the next step?
    int(0topi)1/3edtheta
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  4. #4
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    Quote Originally Posted by chevy900ss View Post
    is the next step?
    int(0topi)1/3edtheta

    \int re^{r^2}dr=\frac{1}{2}e^{r^2}.
    I can't understand what you wrote.

    Tonio
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  5. #5
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    yeah tried to skip steps but, where did the r^2/2 go? I thought it would have been = to ((1/2)r^2)e^r^2
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  6. #6
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    Quote Originally Posted by chevy900ss View Post
    yeah tried to skip steps but, where did the r^2/2 go? I thought it would have been = to ((1/2)r^2)e^r^2

    What kind of question is that? Derivate my solution and check whether you get what's inside of the integral or not!

    tonio
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  7. #7
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    ok ill start over, i didnt mean to sound like i was right and you were wrong. sorry if i came off that way. I really barely have a clue what im doing. i got that the integral of [re^(r^2)]dr equals [(r^2/2)e^(r^2)].
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  8. #8
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    Quote Originally Posted by chevy900ss View Post
    ok ill start over, i didnt mean to sound like i was right and you were wrong. sorry if i came off that way. I really barely have a clue what im doing. i got that the integral of [re^(r^2)]dr equals [(r^2/2)e^(r^2)].

    Well that's false, and checking it is very simple: derivate the "solution" and you'll find out that it doesn't equal the integrand!
    One MUST know derivation very-very well to go into antiderivation/indefinite integration, otherwise things get tough.

    Tonio
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