15. $iiint_e (1/x^{3})dv$, where $e = { (x,y,z) mid 0leq yleq 1,0leq zleq y^{2},1leq xleq z + 1}$

Explanation:

Step1: Set up the triple - integral

The triple - integral $\iiint_E\frac{1}{x^{3}}dV$ with the limits of integration given by $E={(x,y,z)\mid0\leq y\leq1,0\leq z\leq y^{2},1\leq x\leq z + 1}$ can be written as $\int_{0}^{1}\int_{0}^{y^{2}}\int_{1}^{z + 1}\frac{1}{x^{3}}dxdzdy$.

Step2: Integrate with respect to $x$

First, integrate $\int_{1}^{z + 1}\frac{1}{x^{3}}dx$. Using the power - rule for integration $\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\int\frac{1}{x^{3}}dx=-\frac{1}{2x^{2}}+C$. Evaluating the definite integral: [ \begin{align*} \int_{1}^{z + 1}\frac{1}{x^{3}}dx&=-\frac{1}{2(z + 1)^{2}}+\frac{1}{2\times1^{2}}\ &=\frac{1}{2}-\frac{1}{2(z + 1)^{2}} \end{align*} ]

Step3: Integrate the result with respect to $z$

Now, integrate $\int_{0}^{y^{2}}\left(\frac{1}{2}-\frac{1}{2(z + 1)^{2}}\right)dz$. [ \begin{align*} \int_{0}^{y^{2}}\left(\frac{1}{2}-\frac{1}{2(z + 1)^{2}}\right)dz&=\int_{0}^{y^{2}}\frac{1}{2}dz-\frac{1}{2}\int_{0}^{y^{2}}\frac{1}{(z + 1)^{2}}dz\ &=\frac{1}{2}z\big|{0}^{y^{2}}+\frac{1}{2}\frac{1}{z + 1}\big|{0}^{y^{2}}\ &=\frac{y^{2}}{2}+\frac{1}{2(y^{2}+1)}-\frac{1}{2} \end{align*} ]

Step4: Integrate the result with respect to $y$

Finally, integrate $\int_{0}^{1}\left(\frac{y^{2}}{2}+\frac{1}{2(y^{2}+1)}-\frac{1}{2}\right)dy$. [ \begin{align*} \int_{0}^{1}\frac{y^{2}}{2}dy+\frac{1}{2}\int_{0}^{1}\frac{1}{y^{2}+1}dy-\frac{1}{2}\int_{0}^{1}dy&=\frac{1}{6}y^{3}\big|{0}^{1}+\frac{1}{2}\arctan(y)\big|{0}^{1}-\frac{1}{2}y\big|_{0}^{1}\ &=\frac{1}{6}+\frac{\pi}{8}-\frac{1}{2}\ &=\frac{\pi}{8}-\frac{1}{3} \end{align*} ]

Answer:

$\frac{\pi}{8}-\frac{1}{3}$