Changing bounds of double integral while converting from cartesian to polar form

Question

Evaluate the integral $\displaystyle \int_{0}^{\sqrt{π}} \int_{y}^{\sqrt{π}} \cos(x^2) \ \mathrm{d}x \, \mathrm{d}y $

I have replaced $x^2$ with $r^2\cos^2(θ)$ but could not change the bounds by using inequalities.

Answer 1

You are integrating over triangle bound by $y = 0, x=y, x = \sqrt {\pi}$ so changing the order your integral becomes

$\displaystyle \int_{0}^{\sqrt{π}} dx \,\int_{0}^{x} \cos(x^2) \, dy$

Now you can use substitution $x^2 = t$ for your second integral.

Answer 2

Going off of the hint given by azif00 and completing the work done by sumanta we have \begin{align*} \int_0^{\sqrt{\pi}}\int_{x=y}^{x=\sqrt{\pi}} \cos(x^2)dxdy &= \int_0^\sqrt{\pi}\int_{y=0}^{y=x}\cos(x^2) dydx \\ &= \int_0^\sqrt{\pi}\left[y\cos(x^2)\right]_{y=0}^{y=x}dx \\ &= \int_0^\sqrt{\pi}x\cos(x^2)- (0\cdot\cos(x^2)) \ dx \\ &= \int_0^\pi x\cos(u)\frac{du}{2x} \tag{substitute $u = x^2$}\\ &= \frac12\int_0^\pi\cos(u)du \\ &= \frac12\Big[\sin(u)\Big]_0^\pi \\ &= \frac12 (0 - 0) \\ &= 0. \end{align*} Note that it often helps to draw the region over which we are integrating:

Answer 3

If it is obligatory desire to use polar coordinates, then $x=r \cos \phi, y= r \sin \phi$ gives $$\left\{\begin{array}{l} 0 \leqslant x \leqslant \sqrt{\pi} \\ 0 \leqslant y \leqslant x \end{array}\right\} \to \left\{\begin{array}{l} 0 \leqslant r \cos \phi \leqslant \sqrt{\pi} \\ 0 \leqslant r \sin \phi \leqslant r \cos \phi \end{array}\right\}$$

First of equations gives restriction for radius $r \leqslant \frac{\sqrt{\pi} }{\cos \phi}$ and second for angle $\phi \in [0, \frac{\pi}{4}]$, so we have

$$ \int\limits_{0}^{\sqrt{π}} dx \int\limits_{0}^{x} \cos(x^2)dy=\int\limits_{0}^{\frac{\pi}{4}} \int\limits_{0}^{\frac{\sqrt{\pi} }{\cos \phi}}r\cos (r^2 \cos^2 \phi) drd\phi$$

Though, changing order of variables, without polar coordinates, seems to be more good way, but boss is Boss.