Question

Use the method of cylindrical shells to find the volume V of the solid obtained by rotating the region bounded by the given curves about the x-axis. x + y = 7, x = 16 − (y − 3)2

Answer 1

Answer:

$v= \frac{2401\pi}{6}$

Step-by-step explanation:

I put the equations in a plotter, you can see the rotation region below. Now, as we are rotationg about the x-axis we are going to use the equations and the limit points in terms of y. Now, the volume is

$v = 2\pi\int\limits^a_b {p(y)h(y)} \, dy$ where p(y) is the distance from the rotation axis and the differential, a and b are the limit points of the region and h(y) is the height of the region. In this case p(y)= y, a=0, b=7 and h(y) = 16-(y-3)^2-(7-y) (right minus left). So,

$v = 2\pi\int\limits^0_7 {y(16-(y-3)^{2}-7+y)} \, dy$

$v = 2\pi\int\limits^0_7 {y(9-(y-3)^{2}+y)} \, dy$

$v = 2\pi\int\limits^0_7 {y(9-(y^{2}-6y+9)+y)} \, dy$

$v = 2\pi\int\limits^0_7 {y(9-y^{2}+6y-9+y)} \, dy$

$v = 2\pi\int\limits^0_7 {y(-y^{2}+7y)} \, dy$

$v = 2\pi\int\limits^0_7 {-y^{3}+7y^{2}} \, dy$

$v= 2\pi (-\frac{y^{4}}{4}+\frac{7y^{3}}{3})^{7}_0$

$v= 2\pi (-\frac{7^{4}}{4}+\frac{7*7^{3}}{3})$

$v= 2\pi (-\frac{2401}{4}+\frac{2401}{3})$

$v= 2\pi (\frac{2401}{12})$

$v= \frac{2401\pi}{6}$.