Looks familiar?

Let, $\boldsymbol {I=\displaystyle\int_{-4}^{4} y\, dx}$ , where $\boldsymbol {y=\sqrt{16-x^2}}$ .

Now, $\boldsymbol {y=\sqrt{16-x^2}}$ $\boldsymbol {\Rightarrow y^2=16-x^2}$ $\boldsymbol {\Rightarrow x^2+y^2=16}$

It is the equation of a circle with its center at origin $(0,0)$ , which has a radius of 4.

As $\boldsymbol y$ is positive, it will be the semicircle above X -axis and the area of $\boldsymbol {\displaystyle \int_{-4}^{4} y \, dx}$ will be the area of the semicircle. $\boldsymbol {\therefore I=\frac{\pi \times 4^2}{2}=}\color{#69047E} {\boxed {8\pi}}$

The expression can be integrated as follows for $a>0$ , $\displaystyle\int_{-4}^{4} \sqrt{a^{2}-x^{2}} dx=\left [\frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2} \sin^{-1}{\frac{x}{a}}\right]_{-4}^{4}$ where $a=4$ .