How to transform x-y coordinates to polar form?

Question

I have an assignment and I am not sure how to solve this one. Im given this integral below and they said using the transformation $x = rcos\theta$ and $y = rsin\theta$ rewrite the integral below in the $r - \theta$ coordinate system and insert the appropriate limits of integration. $$\int_{-1}^{1}\int_{0}^{\sqrt{1-y^2}}3dxdy$$

How do I go about this or solve this? I am really lost.

Answer 1

It seems to be the integral of the constant function $3$ on the right semi-unit circle. Thus, $r$ varies from $0$ to $1$ and $\theta$ from $-\frac\pi2$ to $\frac\pi2$. The differential elements $\,\mathrm dx\,\mathrm dy$ have to be replaced with $ r\,\mathrm d r\,\mathrm d \theta$, so we obtain $$\int_0^13r\,\mathrm d r\int_{-\frac\pi 2}^{\frac\pi 2}\mathrm d \theta=\frac{3\pi}2$$ which could have been foreseen without any calculation…

Answer 2

The domain is the upper half of a circle centered at the origin with radius 1 then recall that $dxdy=rdrd\theta$ and thus

$$\int_{-1}^{1}\int_{0}^{\sqrt{1-y^2}}3dxdy=\int_{0}^{\pi}d\theta\int_{0}^{1}3rdr$$

Answer 3

Try to draw the region. As far as I can see, it will be a half circle on the right side. So, $r \in [0, 1]$ and $\theta \in [-\pi/2, \pi/2]$. And don't forget the jacobian.