Double Integrals over General Regions — Question 7

Question 7 – Solution

Strategy. Use vertical slices of the triangular base, then compare the integral with the intercept-form tetrahedron.

Step 1: Bounds For $0\le x\le 1$, the variable $y$ runs from $0$ to $1-x$. Therefore $$V=\int_0^1\int_0^{1-x}(1-x-y)\,dy\,dx.$$

Step 2: Evaluate $$\begin{align*} V&=\int_0^1\left[(1-x)y-\frac{y^2}{2}\right]_0^{1-x}dx\\ &=\frac 12\int_0^1(1-x)^2dx =\frac 12\left[x-x^2+\frac{x^3}{3}\right]_0^1\\ &=\boxed{\frac 16}. \end{align*}$$

Step 3: Geometric check The coordinate planes and $x+y+z=1$ enclose a tetrahedron with mutually perpendicular intercept lengths $1,1,1$. Its volume is $$\frac 13(\text{base area})(\text{height}) =\frac 13\left(\frac 12\right)(1)=\frac 16.$$ This matches the integral and verifies the top surface remains nonnegative over $D$.