The value of the integral

\(\int_{-1}^{1} \int_{0}^{\sqrt{1-x^{2}}}\left(x^{2}+y^{2}\right)^{\frac{3}{2}} d y d x\) is:

Concept:

The given integral is: \( \int_{-1}^{1} \int_{0}^{\sqrt{1 - x^2}} (x^2 + y^2)^{\frac{3}{2}} \, dy\, dx \)

This represents a double integral over the upper half of the unit circle centered at the origin. So, converting it into polar coordinates simplifies the integration process.

Let \( x = r\cos\theta,\ y = r\sin\theta \) and \( dx\,dy = r\,dr\,d\theta \)

The region of integration becomes a semicircle: \( r \in [0,1],\ \theta \in [0,\pi] \)

Calculation:

In polar coordinates, \( x^2 + y^2 = r^2 \Rightarrow (x^2 + y^2)^{3/2} = r^3 \)

So, the integral becomes:

\( \int_{0}^{\pi} \int_{0}^{1} r^3 \cdot r \, dr\, d\theta = \int_{0}^{\pi} \int_{0}^{1} r^4 \, dr\, d\theta \)

\(= \int_{0}^{\pi} \left[\frac{r^5}{5}\right]_{0}^{1} d\theta = \int_{0}^{\pi} \frac{1}{5} \, d\theta = \frac{\pi}{5} \)

Correct Option:

The correct answer is: 3) π/5