Question

Evaluate the given integral by changing to polar coordinates. R (4x − y) dA, where R is the region in the first quadrant enclosed by the circle x2 + y2 = 16 and the lines x = 0 and y = x

Answer 1

$x=r\cos t$

$y=r\sin t$

$\mathrm dA=\mathrm dx\,\mathrm dy=r\,\mathrm dr\,\mathrm d\theta$

The region $R$ is given for $0\le r\le4$ and $\dfrac\pi4\le t\le\dfrac\pi2$. So we have

$\displaystyle\iint_R(4x-y)\,\mathrm dA=\int_{\pi/4}^{\pi/2}\int_0^4(4r\cos t-r\sin t)r\,\mathrm dr\,\mathrm dt=\frac{32}3(8-5\sqrt2)$