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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.

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Here is how ...

enter image description here

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.

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    I see, thank you for that illustration. I kept visualizing the triangle being turned 2 dimensionally across the y axis. Which would mean the area would just stay the same, but obviously I was wrong. – 2017-01-19