Question

Evaluate

$\int\int_D(3-x-y)dA$

[dA means dxdy or dydx]

where D is the triangle in the (x, y) plane bounded by the x-axis and the lines y = x

and x = 1.

Answer 1

Solution.

$\iint_D(3-x-y)dxdy=\int^1_0(\int^{y=x}_{y=0}(3-x-y)dy)dx=\int^1_0((3y-xy-\dfrac{y^2}{2})\mid^{y=x}_{y=0})dx=$

$=\int^1_0(3x-x^2-\dfrac{x^2}{2})dx=(3\dfrac{x^2}{2}-\dfrac{x^3}{3}-\dfrac{x^3}{6})\mid^1_0=\dfrac{3}{2}-\dfrac{1}{3}-\dfrac{1}{6}=1.$

Answer: 1.