Question

The integral of \(\int_{-1}^1 \int_0^z \int_{x-z}^{x+z} (x+y+z)\,dy \,dx \,dz\) is given by _______

(a) 0

(b) 1

(c) 0.25

(d) 4

The question was posed to me in an online interview.

The origin of the question is Triple Integral in portion Multiple Integrals of Engineering Mathematics

Answer 1

Right answer is (a) 0

For explanation: \(=\int_{z=-1}^1 \int_{x=0}^z \int_{y=x-z}^{x+z}(x+y+z)dy \,dx \,dz = \int_{z=-1}^1 \int_{x=0}^z\Big[xy + \frac{y^2}{2} + zy\Big]_{y=x-z}^{x+z} \,dx \,dz\)

\( =\int_{z=-1}^1 \int_{x=0}^z \Big\{x((x+z)-(x-z))+\frac{1}{2} [(x+z)^2-(x-z)^2] \\

+z((x+z)-(x- z))\Big\}dx \,dz\)

\( =\int_{z=-1}^1 \int_{x=0}^z(2xz+2xz+2z^2)dx \,dz\)

\( = \int_{-1}^1\big[z(2x^2)+(2z^2 x)\big]_{x=0}^z \,dz = \int_{z=-1}^1 2z^3+2z^3 dz=\big[z^4\big]_{-1}^1=0.\)