Answer to Question #219584 in Differential Equations for Rehema

Question

Find the volume of the solid of revolution by rotating the region formed by y=3-x² and y=2 about the line y=2

Answer 1

The given region in cartesian plan ooks like -

If this is rotated about y=2 , we get -

we have -

For ease of calculation, it is convenient if we shift this solid down so the axis of rotation falls along the X-axis:

Note that the radius (relative to the X-axis) of this shifted volume is

equal to $x^{2}-1$ for $x\in[-1,+1]$

and we can slice this solids into thin disk, each with a thickness of $\delta$ , so that each disc has volume of -

${\delta}prr^{2}=\pi{\delta}(x^{2}-1)^{2}$

With very small values of ${\delta}$

 the sum of the volumes of all such disks will be the volume of the rotated solid.

We can evaluate this sum with ${\delta}$ $\to$ 0

 using the integral:

$=\int_{-1}^{1}{\pi}(x^{2}-1)^{2}dx$

$={\pi}\int_{-1}^{1}x^{4}-2x^{2}+1dx$

$={\pi}(\dfrac{x^5}{5}$ $-2\dfrac{x^{3}}{3}$ $+x)|_{-1}^{1}$

$=$ ${\pi}(\dfrac{3-10+15}{15}-\dfrac{({-3})+10-15}{15})=\dfrac{16}{15}\pi$