# Problems from Changes of variables in double integrals

Compute the integral of the function $$f(x,y)=x^2+y^2$$ on the region between the $$x$$ axes and the $$45$$ degree line, with radius between $$1$$ and $$2$$.

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### Development:

We take the polar coordinates as new variables: $$\begin{array}{l} u(x,y)=r(x,y)=\sqrt{x^2+y^2} \\ v(x,y)=\theta(x,y)=\arctan\Big(\dfrac{y}{x}\Big)\end{array}$$$or similarly: $$\begin{array}{l} x=r\cdot\cos\theta \\ y=r\cdot\sin\theta \end{array}$$$ Then $$J=\begin{bmatrix} \cos\theta & -r\cdot\sin\theta \\ \sin\theta & r\cdot\cos\theta \end{bmatrix}$$ and $$|J|=r$$.

By inspection of the drawing, we can see that the new integration region is $$\begin{array}{c} r\in[1,2] \\ \theta\in[0,\pi/4] \end{array}$$ and $$\widehat{f}(r,\theta)=r^2$$.

Then: $$\int_R (x^2+y^2) \ dxdy = \int_0^{\pi/4}\int_1^2 r^2\cdot r\cdot drd\theta=\int_0^{\pi/4}\int_1^2 r^3drd\theta$$$$$\int_0^{\pi/4} \Big[\dfrac{r^4}{4}\Big]_1^2 \ d\theta=\int_0^{\pi/4} \dfrac{15}{4}d\theta=\dfrac{15}{16}\pi$$$

### Solution:

$$\displaystyle \int_R (x^2+y^2) \ dxdy = \dfrac{15}{16}\pi$$

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