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Math Help - changing variable integral mmn15 1B

  1. #1
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    changing variable integral mmn15 1B

    \int_{0}^{5}\int_{0}^{\sqrt{{25-x^{2}}}}ln(1+x^{2}+y^{2})dxdy
    how to solve it?
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  2. #2
    Ted
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    By changing to Polar Coordinates.
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  3. #3
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    ok x^2+y^2=r
    how do i change dxdy into polar
    how do i find the yacobian of it?
    thetop interval of the inner interval is sqrt(25-r^2cos^2 theta)
    very not simpathetic interval
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  4. #4
    Ted
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    First, it should be dy dx not dx dy in your integral.

    No need for the jocabian here.

    For polar coordinates, dx dy and dy dx = r \, dr \, d\theta

    Your region is D=(x,y) : 0 \leq x \leq 5 and 0 \leq y \leq \sqrt{25-x^2}

    Your integral in polar coordinates is : \int \int_R r \, ln(r^2+1) \, dr \, d\theta

    where R=(r,\theta) : \alpha \leq \theta \leq \beta and a \leq r \leq b

    To find \alpha , \beta , a and  b.., Draw D and convert it to polar coordinates system.
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  5. #5
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    Quote Originally Posted by transgalactic View Post
    ok x^2+y^2=r
    how do i change dxdy into polar
    how do i find the yacobian of it?
    thetop interval of the inner interval is sqrt(25-r^2cos^2 theta)
    very not simpathetic interval
    How in the world did you get that? As you say, r= x^2+ y^2 so sqrt(25- x^2- y^2) is sqrt(25- r^2). There is no "cos^2(theta)". (That would be just 25- x^2.)

    And you should not need to worry about the Jacobian- when you first learned to integrate in polar coordinates you should have learned that dxdy becomes r dr dtheta.

    (That is, of course, because the Jacobian is "r":
    \left|\begin{array}{cc}\frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial \theta} & \frac{\partial y}{\partial \theta}\end{array}\right| = \left|\begin{array}{cc} cos(\theta) & sin(\theta) \\ -r sin(\theta) & r cos(\theta)\end{array}\right|= r cos^2(\theta)+ r sin^2(\theta)= r)
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