Question

Use symmetry to evaluate the double integral \(\iint\limits_R {\frac{{xy}}{{1 + {x^4}}}dA}\), \(R = \{ (x, y)| - 1 \le x \le 1,0 \le y \le 1\} \).

Short answer

Thus, the value of value of integral is \(\iint\limits_R {\frac{{xy}}{{1 + {x^4}}}dA} = 0\)

Step-by-step solution

Factoring out \(y\)

\(\int\limits_0^1 {\int\limits_{ - 1}^1 {\frac{{xy}}{{1 + {x^4}}}dxdy = \int\limits_0^1 y \int\limits_{ - 1}^1 {\frac{x}{{1 + {x^4}}}} } } dxdy\)

Written in this form, it’s clear that inner integralis odd i.e, \(f(x) = - f( - x)\)

If integral is odd,

\(\int\limits_0^1 {\int\limits_{ - 1}^1 {\frac{{xy}}{{1 + {x^4}}}dxdy = \int\limits_0^1 {0{\rm{ }}} } } dy = 0\)

Hence, answer is 0.