Question

Multivariable Calculus

Evaluate the following expression.

Answer 1

Convert to polar coordinates. The domain of integration is a circular arc subtended by a central angle measuring π/4 radians or 45 degrees (the angle
`y=x` makes with the positive
`x`-axis). So we have

`\displaystyle \int_0^(\sqrt2) \int_0^x √(x^2+y^2) \, dy \, dx + \int_(\sqrt2)^2 \int_0^(√(4-x^2)) √(x^2+y^2) \, dy \, dx = \int_0^(\pi/4) \int_0^2 √(r^2) \, r \, dr \, d\theta`

where we use

`\begin{cases} x = r\cos(\theta) \\ y = r\sin(\theta) \\ x^2+y^2 = r^2 \\ dx\,dy = r\,dr\,d\theta\end{cases}`

Computing the integral, we get

`\displaystyle \int_0^(\pi/4) \int_0^2 r^2 \, dr\,d\theta = \frac\pi4 \int_0^2 r^2 \, dr \\\\ ~~~~~~~~ = \frac\pi4 \cdot \frac13 (2^3 - 0^3) = \boxed{\frac{2\pi}3}`