Question

Use symmetry to evaluate the double integral

\(R = ( - \pi ,\pi ) \times ( - \pi ,\pi )\)

Short answer

Thus, the value of the integral is\(4{\pi ^2}\).

Step-by-step solution

Separating the integral

\(\int\limits_{ - \pi }^\pi {\int\limits_{ - \pi }^\pi {1dxdy + } } \int\limits_{ - \pi }^\pi {\int\limits_{ - \pi }^\pi {{x^2}\sin ydxdy + } } \int\limits_{ - \pi }^\pi {\int\limits_{ - \pi }^\pi {{y^2}\sin xdxdy} } \)

Changing the order of integration

\(\begin{array}{c}\int\limits_{ - \pi }^\pi {\int\limits_{ - \pi }^\pi {\left( {1 + {x^2}\sin y + {y^2}\sin xdxdy} \right)} } \\ = \int\limits_{ - \pi }^\pi {\int\limits_{ - \pi }^\pi {1dxdy + } } \int\limits_{ - \pi }^\pi {{x^2}\int\limits_{ - \pi }^\pi {\sin ydydx + } } \int\limits_{ - \pi }^\pi {{y^2}\int\limits_{ - \pi }^\pi {\sin xdxdy} } \end{array}\)

Observing that first integral is a constant& the second two are both oddfunctions over symmetric intervals.

So,

\(\begin{array}{c}\int\limits_{ - \pi }^\pi {\int\limits_{ - \pi }^\pi {\left( {1 + {x^2}\sin y + {y^2}\sin xdxdy} \right)} } \\ = 4{\pi ^2} + \int\limits_{ - \pi }^\pi {0dx} + \int\limits_{ - \pi }^\pi {0dy} \\ = 4{\pi ^2}\end{array}\)

Hence, the answer is \(4{\pi ^2}\).