calculate-D-x-2-y-2-dxdy-with-D-1-1-2-

Question Number 41343 by maxmathsup by imad last updated on 05/Aug/18

c a l c u l a t e \int \int_{D} (x^{2} - y^{2}) d x d y w i t h

D = {[- 1, 1]}^{2}

Commented by maxmathsup by imad last updated on 08/Aug/18

I = \int_{- 1}^{1} (\int_{- 1}^{1} (x^{2} - y^{2}) d x) d y b u t

\int_{- 1}^{1} (x^{2} - y^{2}) d x = 2 \int_{0}^{1} (x^{2} - y^{2}) d x = 2 {[\frac{x^{3}}{3} - y^{2} x]}_{0}^{1}

= 2 {\frac{1}{3} - y^{2}} = \frac{2}{3} - 2 y^{2} \Rightarrow I = \int_{- 1}^{1} (\frac{2}{3} - 2 y^{2}) d y

= \frac{4}{3} - 2 \int_{- 1}^{1} y^{2} d y = \frac{4}{3} - 4 \int_{0}^{1} y^{2} d y = \frac{4}{3} - 4 {[\frac{y^{3}}{3}]}_{0}^{1} = \frac{4}{3} - \frac{4}{3} \Rightarrow

I = 0

Answered by alex041103 last updated on 08/Aug/18

\int \int_{D} (x^{2} - y^{2}) d x d y = \int_{- 1}^{1} \int_{- 1}^{1} (x^{2} - y^{2}) d x d y =

= \int_{- 1}^{1} {[\frac{x^{3}}{3} - y^{2} x]}_{x = - 1}^{x = 1} d y =

= \int_{- 1}^{1} (\frac{2}{3} - 2 y^{2}) d y = {[\frac{2}{3} y - \frac{2 y^{3}}{3}]}_{y = - 1}^{y = 1} =

= \frac{4}{3} - \frac{4}{3} = 0

\int \int_{D} (x^{2} - y^{2}) d x d y = 0