Converting double integral from cartesian to polar coordinates

Question

I want to find $\displaystyle \int_1^4 \int_{0}^{\sqrt{x}}\exp(y/x)\,\mathrm dy\mathrm dx$ by transforming the integral to polar form.

The region of integration is a part of the area under $\sqrt{x} $. By intuition I can say that the lower bound for $r$ is $1$. $\theta$ changes from $0$ to $\pi /4$. However, can you please explain how to write the polar coordinates as clear as you can please?

Answer 1

Consider the line $y = (\tan \theta )x$.

• Let's first figure out the range of $\theta$, clearly, the lower limit is $0$. and the upper limit is when it intersect with the point $(1,1)$, that is when $\theta = \frac{\pi}4$.

• We also want to figure out when does $y=(\tan \theta)x$ intersect with $x=4$ rather than $y=\sqrt{x}$. For that, we study the point $(4,2)$. That is $\tan \theta = \frac12$

• Now let's first the distance from the origin when it intersect with $x=1$. From trigonometry, we have $r\cos \theta = 1$. That is the lower limit of $r$ is $r=\sec \theta$.

• Next, we study the distance from the origin when it intersect with $x=4$. From trigonometry, we have $r\cos \theta = 4$. That is the upper limit of $r$ is $r=4\sec\theta$ if it intersects with $x=4$ as the upper limit.

• If the upper limit of the line intersect with $\sqrt{x}$. The intersection happens at $$\tan \theta=\frac1{\sqrt{x}}$$

$$r^2=x^2+x=\cot^4 \theta +\cot^2 \theta $$

$$\int_1^4\int_0^{\sqrt{x}}e^{\frac{y}{x}}\,\, dydx=\int_0^{\tan^{-1}(0.5)}\int_{\sec\theta}^{4\sec\theta}e^{\tan \theta}r\, drd\theta+\int_{\tan^{-1}(0.5)}^{\frac{\pi}4}\int_{\sec\theta}^{\sqrt{\cot^4 \theta+\cot^2 \theta}}e^{\tan \theta}r\, drd\theta$$

Answer 2

Hint: $$0\le y \le\sqrt x\iff 0\le r\sin\theta\le\sqrt{r\cos\theta}.$$ but the segments $x = 1$, $y\in[0,\cdots]$ and $x = 4$, $y\in[0,\cdots]$ are part of the border.

More: the relevant points for $\theta$ (and $r$) are $(1,0)$, $(4,0)$, $(4,2)$ and $(1,1)$.