Answer to Question #151573 in Calculus for Joshua

Question

Find the volume generated by revolving the area bounded by the lines x+2y=4, x=0 and y=0, about x-axis.

Answer 1

$$x+2y=4=>y=-\dfrac{1}{2}x+2$$ $$y=0: 0=-\dfrac{1}{2}x+2=>x=4, Point (4,0)$$

$$V=\pi\displaystyle\int_{0}^4(-\dfrac{1}{2}x+2)^2dx=\pi\displaystyle\int_{0}^4(\dfrac{1}{4}x^2-2x+4)^2dx$$

$$=\pi\big[\dfrac{1}{12}x^3-x^2+4x\big]\begin{matrix} 4 \\ 0 \end{matrix}=\pi(\dfrac{64}{12}-16+16-0)$$

$$=\dfrac{16\pi}{3} (units^3)$$

$V=\dfrac{16\pi}{3}\ cubic\ units$