How do you find the volume bounded by x-8y=0 & the lines x+2y revolved about the y-axis?

May 21, 2018

Volume of revolution =$\frac{188}{3} \pi {a}^{3}$

Explanation:

$x - 8 y = 0$, PASSES through the origin
$x + 2 y = a$ has the intercepts,
$x = a , \mathmr{and} y = \frac{a}{2}$
The ihtersection of the two lines happen to be,
$8 y + 2 y = a$
$y = \frac{a}{10}$
$x = 8 \frac{a}{10}$

^3Considering y axis to represent the height of the cone,
Area of cross section is given by
$A = \pi {\left(r\right)}^{2}$
where,
r is the x coordinate.
For x-8y=0, from y=0 to y=a/10,
x=8y
A=pi(x)^2
A=pi(8y)^2
$A = 64 \pi {y}^{2}$
Volume1 = $\int A \mathrm{dy} = {\int}_{0}^{\frac{a}{10}} 64 \pi {y}^{2} \mathrm{dy} = 64 \frac{\pi}{3} {y}^{3} = 64 \frac{\pi}{3} \left({\left(\frac{a}{10}\right)}^{3} - {0}^{3}\right)$
$V o l u m e 1 = \frac{64}{1000} \frac{\pi}{3} {a}^{3}$
For x=2y=1 from y=a/10 to y=a/2
$x = a - 2 y$
A=pi(x)^2
$A = \pi {\left(a - 2 y\right)}^{2}$
Volume2=$\int A \mathrm{dy} = {\int}_{\frac{a}{10}}^{\frac{a}{2}} \pi {\left(a - 2 y\right)}^{2} \mathrm{dy}$
=-pi/2(a-2y)^3/3=-pi/6((a-2xx(a/2)^3-(a-2xxa/10)^3)

$= - \frac{\pi}{6} \left(a - \frac{2}{8} {a}^{3} - a + \frac{2}{1000} {a}^{3}\right)$
$= - \frac{\pi}{6} \left(- \frac{2}{8} {a}^{3} + \frac{2}{1000} {a}^{3}\right)$

$= 2 \left(\frac{\pi}{6}\right) \left(\frac{1}{8} - \frac{1}{1000}\right) {a}^{3}$
$\frac{1}{8} = \frac{125}{1000}$
$= \frac{\pi}{3} \left(\frac{125}{1000} - \frac{1}{1000}\right) {a}^{3}$

Volume2=$124 \frac{\pi}{1000} {a}^{3}$

Volume of revolution = Volume1+Volume2
$= \frac{64}{1000} \left(\frac{\pi}{3}\right) {a}^{3} + \frac{124}{1000} \left(\frac{\pi}{3}\right) {a}^{3}$
$64 + 124 = 188$
Volume of revolution =$\frac{188}{3} \pi {a}^{3}$