volume of solid of revolution further integration

May 2015
136
1
United Kingdom
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:
Dec 2014
133
103
USA
For part (b), the sum of the two $\bar{x}$’s do not carry the same weight as you have done.

Instead,

$\bar{x} = \dfrac{\frac{3}{4} \cdot \frac{4\pi}{3} + \frac{3}{2} \cdot \frac{8\pi}{3}}{4\pi} = \dfrac{5}{4}$
 
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Dec 2014
133
103
USA
Part (a) may also be done using cylindrical shells w/r to $y$ ...

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

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

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


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

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

$6\pi \bigg[\dfrac{y^2}{2} - \dfrac{y^3}{6} \bigg]_0^2 = 6\pi \left(\dfrac{2}{3}\right) = 4\pi$
 
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May 2015
136
1
United Kingdom
For part (b), the sum of the two $\bar{x}$’s do not carry the same weight as you have done.

Instead,

$\bar{x} = \dfrac{\frac{3}{4} \cdot \frac{4\pi}{3} + \frac{3}{2} \cdot \frac{8\pi}{3}}{4\pi} = \dfrac{5}{4}$
please can you explain your approach i want to under stand it so i can successfully tackle the question in the future

thank you