This question was asked in an unofficial SAT 2 Math Exam. If the region bounded by the lines $y = -4/3 x + 4$ , $x=0$, and $y=0$ is rotated about the $y$-axis, the volume of the figure formed is?
I simply don't understand how this could become a cone, or how this could become 3-dimensional whatsoever.
Here is how ...
And the volume of this cone is :
$$V=\frac{bh}{3}$$
where $b$ denotes basis area and $h$ is the height. Here :
$$V=\frac{(\pi\times 3^2)\times 4}{3}=12\pi$$
Note the following general formula :
$V=\pi\int_0^af(x)^2\,dx$
which gives the volume of the body generated by rotation of the curve given by $y=f(x)$ where $0\le x\le a$, around the $x$-axis.