Question

What is the volume of the region under the surface z=xy that lies over the triangle with vertices at the origin, (3, 0), and (0, 6)?

Answer 1

The equation of line joining the points $(3,0)$ and $(0,6)$ is given by,

$y-0=\frac{6-0}{0-3}(x-3)$

$y=-2(x-3)$

$y=6-2x$

Thee triangular region with vertices at origin, $(3,0)$ and $(0,6)$ is as shown in the figure below:

Now, the volume of the region under the surface $z=xy$ over the triangular region is evaluated as,

$V=\iint_{R}zdA$

$=\int_{0}^{3}\int_{0}^{6-2x}xydydx$

$=\int_{0}^{3}x[\frac{y^2}{2}]_{0}^{6-2x}dx$

$=\frac{1}{2}\int_{0}^{3}x(6-2x)^2dx$

$=\frac{1}{2}\int_{0}^{3}x(36+4x^2-24x)dx$

$=\int_{0}^{3}x(18+2x^2-12x)dx$

$=\int_{0}^{3}(18x+2x^3-12x^2)dx$

$=[18(\frac{x^2}{2})+2(\frac{x^4}{4})-12(\frac{x^3}{3})]_{0}^{3}$

$=9(9-0)+\frac{1}{2}(81-0)-4(27-0)$

$=81+\frac{81}{2}-108$

$=\frac{81}{2}-27$

$=\frac{27}{2}$

Therefore, the volume of region under the surface $z=xy$ is $V=\frac{27}{2}$ cubic units.