Question

Find the volume of the solid generated by revolving the region enclosed by the triangle with vertices left parenthesis 4 comma 3 right parenthesis(4,3)​, left parenthesis 4 comma 4 right parenthesis(4,4)​, and left parenthesis 7 comma 4 right parenthesis(7,4) about the​ y-axis.

Answer 1

So we have three points A(4,3), B(4,4) and C(7,4). 
We introduce a fourth point D(7,3).
We obtain a rectangle ABCD. By rotating the rectangle, 
around the y axis, we get two cylinders. Denote the 
first cylinder C1 and the inner cylinder with vertices AB by C2. 
Computing the volume of C1:
$\pi r^2h=\pi7^2\times1=49\pi$
Computing the volume of the cylinder C2:
 $\pi r_2^2h=\pi4^2\times1=16\pi\text{ here the height is 1 again and the radius 4}$
In order to find the volume of the hollow tube, simply compute the difference of the two obtained values like this:
$49\pi-16\pi=33\pi$
Now, in order to find the volume of the solid generated by revolving the region enclosed by the triangle, simply divide the previous value over two like this:
$\frac{33\pi}{2}$

Answer $\frac{33\pi}{2}$

Answer 2

From the sketch of the required triangle, the volume of the solid generated by revolving the region enclosed by the triangle with vertices (4, 3)​, (4, 4)​, and (7, 4) about the​ y-axis is given by

$\pi \int\limits^4_3 {\left[(y-0)^2-(4-0)^2\right]} \, dy =\pi \int\limits^4_3 {\left(y^2-16\right)} \, dy \\ \\ =\pi\left[ \frac{y^3}{3} -16y\right]_3^4=\pi\left[\left( \frac{4^3}{3} -16(4)\right)-\left( \frac{3^3}{3} -16(3)\right)\right] \\ \\ =\pi\left[ \frac{64}{3} -64-9+48\right]= \frac{11\pi}{3}$