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

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[ (y^3)/(3) -16y\right]_3^4=\pi\left[\left( (4^3)/(3) -16(4)\right)-\left( (3^3)/(3) -16(3)\right)\right] \\ \\ =\pi\left[ (64)/(3) -64-9+48\right]= (11\pi)/(3)`

Answer 2

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*1=49\pi`
Computing the volume of the cylinder C2:

`\pi r_2^2h=\pi4^2*1=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:

`(33\pi)/(2)`

Answer
`(33\pi)/(2)`