Double integral, solve by inspection ∫∫$_T (1-x-y) dA$ where $T$ is the triangle with vertices $(0,0), (0,1)$ and $(1,0)$.
I got the answer $1/2$, since $dA=1$ and the area of the triangle is $(1 \cdot 1)/2 = 1/2$. But it's not correct. What do I do wrong?
 A: The issue is that you are integrating $f(x,y)=1-x-y$ not $f(x,y)=1$ which is the z-coordinate if you think about it in the x-y-z plane. You have a two dimensional integral here, not 1D. The way you have approached the problem is by finding the area of a triangle in the x-y plane, you want to find the area of a function in the z-plane with a pyramid base and the top as the function $f(x,y) = 1-x-y$. What you want is, for $f(x,y)=1-x-y$
$\int_T f(x,y)dA = \int_0^1 \int_0^{1-y}(1-x-y)dxdy = \int_0^1 \left[x-\frac{1}{2}x^2 - yx\right]_0^{1-y} dy = \frac{1}{6}$
(I assume you can integrate it from here, so I do not show all of the steps). Also note that you can do the integral by integrating over $y$ first and removing the y-dependence. I.e.
$\int_T f(x,y)dA = \int_0^1 \int_0^{1-x}(1-x-y)dydx = \int_0^1 \left[y-\frac{1}{2}y^2 - yx\right]_0^{1-x} dx = \frac{1}{6}$
A: It's 1/6.
The double integral represents the volume of a pyramid with a triangular base (of area 1/2) and (perpendicular) height of 1.
So the volume is (1/3)(1/2)(1) = 1/6.  (One-third the product of the area of the base and the perpendicular height.)
The "by inspection" part is recognizing the solid shape, given the integration area and the function.  The function value is 1 when (x, y) = (0, 0), and 0 all along the hypotenuse of the isosceles right triangle which bounds the integration area.
