Question

$\text{A triangle in the xy plane is bounded by the straight lines}3x=4y,y=0\text{and}x=4\text{. The volume above the triangle and under the plane}x+y+z=8\text{is}$

• A. 26

Answer 1

The correct option is A 26

Volume=$\int \int \int dzdxdy=\int \int zdydx$

$={\int }_0^4{\int }_0^{\frac{3x}{4}}(8-x-y)dydx={\int }_0^4(8y-xy-\frac{y^2}{2}){|}_0^{\frac{3x}{4}}dx$

$={\int }_0^4(6x-\frac{33x^2}{32})dx=(\frac{6x^2}{2}-\frac{33}{32}\times \frac{x^3}{3}){|}_0^4$

$=(3x^2-\frac{11x^3}{32}){|}_0^4=48-22=26$