# volume of solid of revolution further integration

#### bigmansouf

Question: The triangle formed by the three straight lines y =0, y=2x and y = 3-x is rotated about the side y = 0.
Find the volume of the solid so generated, and the x-coordinate of its centre of gravity.
my attempt:
a) find the volume of the solid
element of the volume = $$\displaystyle \pi y^2 \mathrm{d} x$$
$$\displaystyle v_{1} = \int_{0}^{1}\pi y^2 \mathrm{d} x$$
$$\displaystyle v_{1} = \int_{0}^{1}\pi (2x)^2 \mathrm{d} x = \frac{4}{3} \pi$$

$$\displaystyle v_{2} = \int_{1}^{3}\pi y^2 \mathrm{d} x$$
$$\displaystyle v_{1} = \int_{1}^{3}\pi (x-3)^2 \mathrm{d} x = \frac{8}{3} \pi$$
$$\displaystyle v_{1}+ v_{2} = 4 \pi$$

b) element of volume = $$\displaystyle x y^2 \pi \mathrm{d} x [\tex] \(\displaystyle \bar{x}_{1}\int_{0}^{1} \pi y^2 \mathrm{d} x = \int_{0}^{1} \pi x y^2 \mathrm{d} x$$
$$\displaystyle \bar{x}_{1}\int_{0}^{1} \pi (2x)^2 \mathrm{d} x = \int_{0}^{1} \pi x (2x)^2 \mathrm{d} x$$
$$\displaystyle \bar{x}_{1}\int_{0}^{1} \pi 4x^2 \mathrm{d} x = \int_{0}^{1} \pi 4x^3 \mathrm{d} x$$
$$\displaystyle \bar{x}_{1}\left ( \frac{4}{3} \right ) = 1$$
$$\displaystyle \bar{x}_{1} = \frac{3}{4}$$

$$\displaystyle \bar{x}_{2}\int_{1}^{3} \pi y^2 \mathrm{d} x = \int_{1}^{3} \pi x y^2 \mathrm{d} x$$
$$\displaystyle \bar{x}_{2}\int_{1}^{3} \pi (3-x)^2 \mathrm{d} x = \int_{2}^{3} \pi x (3-x)^2 \mathrm{d} x$$
$$\displaystyle \bar{x}_{2}\int_{1}^{3} \pi (9-6x+x^2) \mathrm{d} x = \int_{1}^{3} \pi (9x-6x^2+x^3) \mathrm{d} x$$
$$\displaystyle \bar{x}_{2}\left ( \frac{8}{3} \right ) = 4$$
$$\displaystyle \bar{x}_{2} = \frac{3}{2}$$
$$\displaystyle \bar{x}_{1} + \bar{x}_{2} = \frac{3}{2} + \frac{3}{4} = \frac{9}{4}$$
the book gives an answer for part a) $$\displaystyle 4 \pi$$ b) $$\displaystyle \frac{5}{4}$$
I applied the same method i used in part a if part b
I want to know if there is another way to do part a
Please can someone help me with part b\)

Last edited:

#### Cervesa

For part (b), the sum of the two $\bar{x}$’s do not carry the same weight as you have done.

$\bar{x} = \dfrac{\frac{3}{4} \cdot \frac{4\pi}{3} + \frac{3}{2} \cdot \frac{8\pi}{3}}{4\pi} = \dfrac{5}{4}$

1 person

#### Cervesa

Part (a) may also be done using cylindrical shells w/r to $y$ ...

$y=3-x \implies x=3-y$

$y=2x \implies x = \dfrac{y}{2}$

$2x=3-x \implies x =1 \implies y=2$

$\displaystyle 2\pi \int_0^2 y \bigg[(3-y)-\dfrac{y}{2}\bigg] \, dy$

$\displaystyle 6\pi \int_0^2 y - \dfrac{y^2}{2} \, dy$

$6\pi \bigg[\dfrac{y^2}{2} - \dfrac{y^3}{6} \bigg]_0^2 = 6\pi \left(\dfrac{2}{3}\right) = 4\pi$

1 person

#### bigmansouf

For part (b), the sum of the two $\bar{x}$’s do not carry the same weight as you have done.

$\bar{x} = \dfrac{\frac{3}{4} \cdot \frac{4\pi}{3} + \frac{3}{2} \cdot \frac{8\pi}{3}}{4\pi} = \dfrac{5}{4}$