# Converting double integral from cartesian to polar coordinates

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?

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 $$\tan \theta=\frac1{\sqrt{x}}$$

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

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

• Hi, why do we consider this line? – Ninja Apr 7 '19 at 18:00
• this line means fixing $\theta$, find the relevant range of values for $r$. – Siong Thye Goh Apr 7 '19 at 18:02
• Alternatively, you can consider circular stripe of distrance $r$ from origin and find the interval of the limit, I haven't explored which is easier. – Siong Thye Goh Apr 7 '19 at 18:21

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 $$(1,1)$$.