Volume of Rotation of a Square

Question

I have a square with vertices of $(0,5), (0,7), (-1, 6) $ and $ (1,6)$. This is to be revolved around the x axis to find the volume of rotation. How do I go about starting this problem?

Thanks in advance!

Answer 1

I think the easiest way to do this in terms of algebra would be to do integration with respect to $y$ and then double the result (you're still rotating it about the x-axis). Your curves are $x=-y+7$ and $x=y-5$. The method used is the shell method:

$$ V=2\left[2\pi\int_{6}^{7}y(7-y)\,dy+2\pi\int_{5}^{6}y(y-5)\,dy\right]=\\ 4\pi\left[\int_{6}^{7}(7y-y^2)\,dy+\int_{5}^{6}(y^2-5y)\,dy\right]=24\pi\ cubic\ units. $$

Answer 2

If you break that square into two triangles, the volume of rotation of the one in the second quadrant is$$\pi\int_{-1}^{0}(x+7)^2-(5-x)^2\,\text{d}x=12\pi.$$