# Thread: The double integral of this problem.

1. ## The double integral of this problem.

Evaluate the iterated integral. Ignore the answer. I want to know how to integrate this. I know you first integrate from respect with x first. But how do you integrate the first part? Trig Substitution? Its been awhile since I done integrals.

2. EDIT: $\int^{3}_{1} \int_{0}^{y} \frac{4}{x^{2}+y^{2}} \ dx \ dy = 4 \int^{3}_{1} \int^{y}_{0} \frac{1}{y^{2}(1+\frac{x^{2}}{y^{2}})} \ dx \ dy$

$= 4 \int^{3}_{1} \frac{1}{y^{2}} \int^{y}_{0} \frac{1}{1+(\frac{x}{y})^2} \ dx dy = 4 \int^{3}_{1} \frac{1}{y} \int_{0}^{1} \frac{1}{1+u^{2}} \ du \ dy =$ $= 4 \int^{3}_{1} \frac{1}{y} \arctan u \Big|^{u=1}_{u=0} \ dy$

$= \pi \int^{3}_{1} \frac{dy}{y} = \pi \ln y \Big|^{3}_{1} = \pi \ln 3$

3. Why did the y change to 3 then to 1?

4. or use polar coordinates:

$\int_{1}^{3}{\int_{0}^{y}{\frac{4}{x^{2}+y^{2}}\,d x}\,dy}=4\int_{0}^{\frac{\pi }{4}}{\int_{\sec \varphi }^{3\sec \varphi }{\frac{dr\,d\varphi}{r}}}=4\ln (3)\int_{0}^{\frac{\pi }{4}}{d\varphi }=\pi \ln 3.$

5. Originally Posted by asdf122345
Why did the y change to 3 then to 1?
Because I made a mistake.