Answer to Question #104620 in Calculus for Gayatri Yadav

Question

The function f(x,y,z) = e^(xyz) is integrable over [0,2] × [0,2] × [0,2] .

Answer 1

We first calculate the integral

$\int\limits_{0}^{2}e^{xyz} dz=\\ =\left\{\begin{matrix} \frac{e^{xyz}}{xy}|_0^2, for xy\neq 0 \\ z|_0^2, otherwise \end{matrix}\right.=\\ \left\{\begin{matrix} \frac{e^{2xy}}{xy}-\frac{1}{xy}, for xy\neq 0 \\ 2, otherwise \end{matrix}\right. .\\$

Then find

$\int\limits_{0}^{2} \left\{\begin{matrix} \frac{e^{2xy}}{xy}-\frac{1}{xy}, for xy\neq 0 \\ 2, otherwise \end{matrix}\right. dy=\\ =\left\{\begin{matrix} \int\limits_{0}^{2}\frac{e^{2xy}-1}{xy}dy, for xy\neq 0 \\ \int\limits_{0}^{2} 2dy, otherwise \end{matrix}\right. =\\ =\left\{\begin{matrix} \int\limits_{0}^{2}\frac{e^{2xy}-1}{xy}dy, for xy\neq 0 \\ 2y|^2_0, otherwise \end{matrix}\right. =\\ =\left\{\begin{matrix} \int\limits_{0}^{2}\frac{e^{2xy}-1}{xy}dy, for xy\neq 0 \\ 4, otherwise \end{matrix}\right. .$

Then find

$\int\limits_{0}^{2}\left\{\begin{matrix} \int\limits_{0}^{2}\frac{e^{2xy}-1}{xy}dy, for xy\neq 0 \\ 4, otherwise \end{matrix}\right. dx=\\ =\left\{\begin{matrix} \int\limits_{0}^{2} \int\limits_{0}^{2}\frac{e^{2xy}-1}{xy}dxdy, for xy\neq 0 \\ \int\limits_{0}^{2} 4 dx, otherwise \end{matrix}\right. =\\ =\left\{\begin{matrix} \int\limits_{0}^{2} \int\limits_{0}^{2}\frac{e^{2xy}-1}{xy}dxdy, for xy\neq 0 \\ 4 x|_0^2, otherwise \end{matrix}\right. =\\ =\left\{\begin{matrix} \int\limits_{0}^{2} \int\limits_{0}^{2}\frac{e^{2xy}-1}{xy}dxdy, for xy\neq 0 \\ 8, otherwise \end{matrix}\right. .$

Consider the function $\frac{e^{2xy}-1}{xy}$ and find limit

$\lim\limits_{xy\mapsto 0} \frac{e^{2xy}-1}{xy}=2$

because

$\lim\limits_{t\mapsto 0} \frac{e^{t}-1}{t}=1$ .

This means that the function is integrable and the integral is convergent.