Question

Evaluate the double integral ∬R(3x−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, by changing to polar coordinates.

Answer 1

`\displaystyle 64-32√(2)+(32√(2))/(3)\approx3.66`

Explanation:

`\displaystyle \iint_R(3x-y)\,dA\\\\=\int^(\pi)/(2)_(\pi)/(4)\int^4_0(3r\cos\theta-r\sin\theta)\,r\,dr\,d\theta\\\\=\int^(\pi)/(2)_(\pi)/(4)\int^4_0(3r^2\cos\theta-r^2\sin\theta)\,dr\,d\theta\\\\=\int^(\pi)/(2)_(\pi)/(4)\int^4_0r^2(3\cos\theta-\sin\theta)\,dr\,d\theta\\\\=\int^(\pi)/(2)_(\pi)/(4)(64)/(3)(3\cos\theta-\sin\theta)\,d\theta\\\\=\int^(\pi)/(2)_(\pi)/(4)\biggr(64\cos\theta-(64)/(3)\sin\theta\biggr)\,d\theta`

`\displaystyle =\biggr(64\sin\theta+(64)/(3)\cos\theta\biggr)\biggr|^(\pi)/(2)_(\pi)/(4)\\\\=\biggr(64\sin(\pi)/(2)+(64)/(3)\cos(\pi)/(2)\biggr)-\biggr(64\sin(\pi)/(4)+(64)/(3)\cos(\pi)/(4)\biggr)\\\\=64-\biggr(64\cdot{(√(2))/(2)}+(64)/(3)\cdot{(√(2))/(2)}\biggr)\\\\=64-32√(2)+(32√(2))/(3)\biggr\\\\\approx3.66`