# Thread: changing variable integral mmn15 1B

1. ## 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?

2. By changing to Polar Coordinates.

3. 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

4. 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.

5. Originally Posted by transgalactic
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":
$\left|\begin{array}{cc}\frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} \\ \frac{\partial x}{\partial \theta} & \frac{\partial y}{\partial \theta}\end{array}\right|$ $= \left|\begin{array}{cc} cos(\theta) & sin(\theta) \\ -r sin(\theta) & r cos(\theta)\end{array}\right|= r cos^2(\theta)+ r sin^2(\theta)= r$)