Definite integral of \(\displaystyle\sqrt{{{16}-{x}^{{2}}}}\) between -4 and 0

Okay, you are looking at half of a circle.

Diameter is 4. So radius is 2.

You are finding an area, the formula is \(\displaystyle{A}=\pi\cdot{r}^{{2}}\)

\(\displaystyle{A}=\pi{2}^{{2}}={4}\pi\)

Okay, you are looking at half of a circle.

Diameter is 4. So radius is 2.

You are finding an area, the formula is \(\displaystyle{A}=\pi\cdot{r}^{{2}}\)

\(\displaystyle{A}=\pi{2}^{{2}}={4}\pi\)