Volume of Revolution for Curve y=(12/(x + 3))- 4 Rotated about the y-axis

  • Thread starter Thread starter jack1234
  • Start date Start date
  • Tags Tags
    Curve Volume
jack1234
Messages
132
Reaction score
0
The question is

Given a curve C whose equation is give by y=(12/(x + 3))- 4,
find the region bounded by C, the line x=0, the line x=3 and the line y=8 is rotated through 360 degree about the y-axis. Calculate the exact value of the volume generated.

My solution is here
http://www.geocities.com/myjunkmail31/Volume.jpg

However, this solution is not the answer in the textbook. The anwser in the textbook is
36pi(1+2ln2).
Can someone show me where am I doing wrong?
 
Last edited:
Physics news on Phys.org
The error is in V1: Easy fix: V_1 = \pi 3^2 (2) - \pi (54 - 2 ln 2).

The volume you calculated is that of the solid generated by rotating the region between the given curve and the y-axis about the y-axis, but you wanted to rotate the region between the given curve and x=3 about the y-axis. So just subtract what you got for V1 from the volume of the cylinder of radius 3 with height 2.
 
Awesome, very thanks=)
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top