Question

Multivariable Calculus - Double Integrals
Evaluate the following equation: ​

Answer 1

The domain of integration is a circular sector subtended by an angle of 3π/4 radians, belonging to a circle of radius 3.

The first integral is taken over the left half of the upper semicircle.
`\left(\frac\pi2 \le \theta \le \pi\right)`

The second integral is taken over a sector belonging to the right half of the same semicircle. Notice that the line
`y=x` meets the semicircle
`y=√(9-x^2)` when

`x = √(9-x^2) \implies x^2 = 9-x^2 \implies x = \frac3{\sqrt2}`

and the line
`y=x` makes an angle of π/4 with the positive
`x`-axis.
`\left(\frac\pi4 \le \theta \le \frac\pi2\right)`

Then the two integrals combine into one integral in polar coordinates, and

`\displaystyle \int_(-3)^0 \int_0^(√(9-x^2)) √(x^2+y^2) \, dy \, dx + \int_0^(3/\sqrt2) \int_0^(√(9-x^2)) √(x^2+y^2) \, dy \, dx \\\\ ~~~~~~~~ = \int_(\pi/4)^\pi \int_0^3 √(r^2)\,r\,dr\,d\theta \\\\ ~~~~~~~~ = \frac{3\pi}4 \int_0^3 r^2 \, dr = \boxed{\frac{27\pi}4}`