Q. 29 - Calculus

Question

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

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

Step-by-Step Solution

Answer

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

Step 1: Given

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

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

Step 2: Evaluate the integral

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