Q. 18 - Calculus

Question

Complete Example $6$ by evaluating the iterated integral ${\int }_{-2}^2{\int }_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}{\int }_0^{x+z+4} ky dy dz dx.$

Step-by-Step Solution

Answer

The value of the iterated integral ${\int }_{-2}^2{\int }_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}{\int }_0^{x+z+4} ky dy dz dx$is,$110k$.

Step 1 . Given information

${\int }_{-2}^2{\int }_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}{\int }_0^{x+z+4} ky dy dz dx.$

Step 2 . Evaluate the given triple integral.

$$\begin{gathered} {\int }_{-2}^2{\int }_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}{\int }_0^{x+z+4} ky dy dz dx={\int }_{-2}^2{\int }_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\left({\int }_0^{x+z+4} kydy\right)dzdx \\ ={\int }_{-2}^2{\int }_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\left(k{\left[\frac{y^2}{2}\right]}_0^{x+z+4}\right)dzdx \\ =\frac{k}{2}{\int }_{-2}^2{\int }_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}{\left(x+z+4\right)}^{2}dzdx \end{gathered}$$

Now expand the obtained expression.

${\int }_{-2}^2{\int }_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}{\int }_0^{x+z+4} ky dy dz dx=\frac{k}{2}{\int }_{-2}^2{\int }_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \left(x^2+z^2+16+2xz+8z+8x\right)dzdx$

Step 3 . Now integrate the obtained expression with respect to z .

$$\begin{gathered} {\int }_{-2}^2{\int }_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}{\int }_0^{x+z+4} ky dy dz dx=\frac{k}{2}{\int }_{-2}^2\left[x^2{\left[z\right]}_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}+{\left[\frac{z^3}{3}\right]}_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}+16{\left[z\right]}_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}+2x{\left[\frac{z^2}{2}\right]}_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}+8{\left[\frac{z^2}{2}\right]}_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}+8x{\left[z\right]}_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\right] \\ = \frac{k}{2}{\int }_{-2}^2\left[x^2\left(2\sqrt{4-x^2}\right)+\frac{2}{3}{\left(\sqrt{4-x^2}\right)}^3+16\left(2\sqrt{4-x^2}\right)+x\left(0\right)+4\left(0\right)+8x\left(2\sqrt{4-x^2}\right)\right]dx \\ =\frac{k}{2}{\int }_{-2}^2\left[2x^2\left(\sqrt{4-x^2}\right)+\frac{2}{3}{\left(4-x^2\right)}^{\frac{3}{2}}+32\left(\sqrt{4-x^2}\right)+16x\left(\sqrt{4-x^2}\right)\right]dx \end{gathered}$$

Step 4 . Now integrate with respect to x .

$$\begin{gathered} {\int }_{-2}^2{\int }_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}{\int }_0^{x+z+4} ky dy dz dx=\frac{k}{2}\left[2{\int }_{-2}^2{x}^2\left(\sqrt{4-x^2}\right)dx+\frac{2}{3}{\int }_{-2}^2{\left(4-x^2\right)}^{\frac{3}{2}dx}+32{\int }_{-2}^2\left(\sqrt{4-x^2}\right)dx+16{\int }_{-2}^{2}x\left(\sqrt{4-x^2}\right)dx\right] \\ =\frac{k}{2}\left[2\mathrm{\pi }+\frac{2}{3}\left(6\mathrm{\pi }\right)+32\left(2\mathrm{\pi }\right)+0\right] \\ =\frac{\mathrm{k}}{2}\left[2\mathrm{\pi }+4\mathrm{\pi }+64\mathrm{\pi }\right] \\ =35\mathrm{k\pi } \\ =110\mathrm{k} \end{gathered}$$

Therefore, the required value is, $110k$.