How do we calculate $ \iint_{D} (1-x-y)dxdy$ using polar coordinates

integration

I have to calculate the following integral in polar coordinates:
$$
\iint\limits_D (1-x-y)dxdy=?
$$
where $D: x^2 + y^2 ⩽ R^2$.
I got the following
$$
\int_0^{2\pi} d\theta\int_0^{R} (1 – rcos\theta – rsin\theta)rdr
$$
My answer is $πR^2$ but I want to confirm if that's correct.

Best Answer

Observe that such integral equals $\pi R^{2}$ (as you have suggested).

That is because the integral of $\sin$ or $\cos$ over a period equals zero.

Consequently, there is no need to evaluate the terms $r^{2}\cos(\theta)$ and $r^{2}\sin(\theta)$.

Hence it remains to calculate the integral: \begin{align*} I = \int_{0}^{2\pi}\mathrm{d}\theta\int_{0}^{R}r\mathrm{d}r \end{align*}

which leads to the desired result.

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