Question

The value of \(\int_0^1 \int_0^x \int_0^{x+y} \,xyz \,dz\, dy\, dx\,\) is given by ____

(a) 17/144

(b) 16/72

(c) 17/72

(d) 15/144

This question was addressed to me at a job interview.

This intriguing question originated from Triple Integral in section Multiple Integrals of Engineering Mathematics

Answer 1

Right option is (a) 17/144

Explanation: \(\int_{x=0}^1 \int_{y=0}^x \int_{z=0}^{x+y}\,xyz \,dz\, dy\, dx\,\)

\(\int_{x=0}^1 \int_{y=0}^x xy[\frac{z^2}{2}]_0^{x+y} \,dy \,dx = \int_{x=0}^1 \int_{y=0}^x xy \frac{(x+y)^2}{2} \,dy \,dx\)

\(=\frac{1}{2} \int_{x=0}^1 \int_{y=0}^x xy(x^2 + y^2 + 2xy) \,dy \,dx\)

\(=\frac{1}{2} \int_{x=0}^1 \int_{y=0}^x(x^3 y + xy^3 + 2x^2 y^2) \,dy \,dx\)

\(=\frac{1}{2} \int_{x=0}^1 \Big[x^3  \frac{y^2}{2} + x \frac{y^4}{4} + \frac{2x^2 y^3}{3}\Big]_{y=0}^x dx = \frac{1}{2} \int_{x=0}^1(\frac{x^5}{2} + \frac{x^5}{4} + \frac{2x^5}{3}) \,dx\)

\(=\frac{1}{2} \Big[\frac{x^6}{12} + \frac{x^6}{24} + \frac{x^6}{9}\Big]_{x=0}^1=(\frac{1}{12} + \frac{1}{24} + \frac{1}{9})\frac{1}{2} = \frac{17}{144}.\)