Question

Evaluate the double integral I=∫∫D xy dA where D is the triangular region with vertices (0,0),(6,0),(0,1).

Answer 1

$S=\iint_D xy dA, where ~ D: 0 \leq x \leq 6, 0 \leq y \leq 1-\frac{x}{6}$

$\iint_D xy dA=\int_0^6 dx \int_0 ^{1-\frac{x}{6}}xydy=\frac{1}{2}\int_0^6x\cdot (1-\frac{x}{6})^2dx=$

$=\frac{1}{2}\int_0^6(x-\frac{x^2}{3}+\frac{x^3}{36})dx=\frac{1}{2}(\frac{x^2}{2}-\frac{x^3}{9}+\frac{x^4}{144})|_0^6=$

$=\frac{1}{2}(18-24+9)=1.5\implies S=1.5$

$Answer:S=1.5$