Calculate
∫(xsinx + y sin(x+y)) dxdy
T
being T = traingle given by the points (1,0) (0,1) (3,3)
How is it that no one responded to this question? For the benefit of other users, here's a response, just over 8 months late, hehe
begin by graphing the region (see below)
the line connecting (0,1) and (1,0) is $\displaystyle y = 1 - x$
the line connecting (1,0) and (3,3) is $\displaystyle y = \frac 32 (x - 1)$
the line connecting (0,1) and (3,3) is $\displaystyle y = \frac 23x + 1$
clearly we have to split this region into two parts: i will split it up into two intervals, [0,1] and [1,3].
Thus, $\displaystyle \iint_T (x \sin x + y \sin (x + y))~dxdy = \iint_T (x \sin x + y \sin (x + y))~dydx = $ $\displaystyle \int_0^1 \int_{1 - x}^{(2/3)x + 1} (x \sin x + y \sin (x + y))~dydx + \int_1^3 \int_{(3/2)(x - 1)}^{(2/3)x + 1} (x \sin x + y \sin (x + y))~dydx$
And continue ... what? ou expected me to give the whole solution away?