Question

Evaluating triple integrals: Each of the triple integrals that follows represents the volume of a solid. Sketch the solid and evaluate the integral.

${\int }_{-2}^2{\int }_0^{\sqrt{4-x^2}}{\int }_0^{y}dzdydx$

Short answer

${\int }_{-2}^2{\int }_0^{\sqrt{4-x^2}}{\int }_0^{y}dzdydx=\frac{16}{3}$

Step-by-step solution

Given

We have to find the volume by using the triple integral.

${\int }_{-2}^2{\int }_0^{\sqrt{4-x^2}}{\int }_0^{y}dzdydx$

Evaluate the integral

$$\begin{gathered} V={\int }_{-2}^2{\int }_0^{\sqrt{4-x^2}}\left[{\int }_0^{y}dz\right]dydx \\ V={\int }_{-2}^2\left[{\int }_0^{\sqrt{4-x^2}}ydy\right]dx \\ V={\int }_{-2}^2{\left[\frac{y^2}{2}\right]}_0^{\sqrt{4-x^2}}dx \\ V=\frac{1}{2}{\int }_{-2}^2(4-x^2)dx \\ V=2\times \frac{1}{2}{\int }_0^2(4-x^2)dx \\ V={\left[4x-\frac{x^3}{3}\right]}_0^2 \\ V=\left[8-\frac{8}{3}\right] \\ V=\frac{16}{3}\mathrm{cubic}\mathrm{units} \end{gathered}$$