Question

The value of the integral
\[\int_{0}^{\sqrt{8}} \int_{x^2}^{\sqrt{8 - x^2}} \frac{3\sqrt{x^2 + y^2}}{\sqrt{8}} \, dy \, dx \]
is equal to __________ (answer in integer).

Answer 1

Correct Answer: 2

Step 1: Rewrite the integral in polar coordinates. The given limits suggest a region bounded by \( x^2 + y^2 \leq 8 \) in the first quadrant. This can be represented in polar coordinates: \[ x = r\cos\theta, \quad y = r\sin\theta, \quad \text{and } x^2 + y^2 = r^2. \] The Jacobian for the transformation is \( r \). The integral in polar coordinates becomes: \[ \int_0^{\pi/4} \int_0^{\sqrt{8}} \frac{3r}{\sqrt{8\pi}} \cdot r \, dr \, d\theta. \] Step 2: Simplify the integral. The integrand becomes: \[ \int_0^{\pi/4} \int_0^{\sqrt{8}} \frac{3r^2}{\sqrt{8\pi}} \, dr \, d\theta. \] First, evaluate the inner integral with respect to \( r \): \[ \int_0^{\sqrt{8}} r^2 \, dr = \left[\frac{r^3}{3}\right]_0^{\sqrt{8}} = \frac{(\sqrt{8})^3}{3} = \frac{8\sqrt{8}}{3}. \] Substitute this into the integral: \[ \int_0^{\pi/4} \frac{3}{\sqrt{8\pi}} \cdot \frac{8\sqrt{8}}{3} \, d\theta = \int_0^{\pi/4} \frac{8}{\sqrt{\pi}} \, d\theta. \] Step 3: Evaluate the remaining integral. The integral with respect to \( \theta \) is: \[ \int_0^{\pi/4} \frac{8}{\sqrt{\pi}} \, d\theta = \frac{8}{\sqrt{\pi}} \cdot \left[\theta\right]_0^{\pi/4} = \frac{8}{\sqrt{\pi}} \cdot \frac{\pi}{4} = 2. \]