Question

A triangle in the $xy-$ plane is bounded by the straight lines $2x=3y,y=0$ and $x=3$. The volume above the triangle and under the plane $x+y+z=6$ is

• A. 10

Answer 1

The correct option is A 10
The triangle in $XY$ plane bounded by lines $y=\frac{2}{3}x,x=3$ and $y=0$ is as shown.
Equation of plane: $x+y+z=6$
$Z=(6-x-y)$ so $0\le z\le 6-x-y$
Now, on $xy$ plane, $0\le y\le \frac{2}{3}x$ and $0\le x\le 3$
so, $V=\int \int \int \left(1\right)dzdydx$
$\therefore V={\int }_{x=0}^3{\int }_{y=0}^{2x/3}(6-x-y)dydx$
$={\int }_{x=0}^3{[6y-xy-\frac{y^2}{2}]}_0^{2x/3}dx$
$={\int }_{x=0}^3(6\times \frac{2x}{3}-x\times \frac{2x}{3}-\frac{4x^2}{9\times 2})dx$
$={\int }_{x=0}^3(4x-\frac{2}{3}{x}^2-\frac{2}{9}{x}^2)dx$
$={[\frac{4x^2}{2}-\frac{2}{3}\frac{x^3}{3}-\frac{2}{9}.\frac{x^3}{3}]}_0^3$
$=\frac{4\times 9}{2}-\frac{2}{9}\times 27-\frac{2}{27}\times 27$
$=10$