# Integration of a volume

• Jul 25th 2013, 03:07 PM
togo
Integration of a volume
Hi, here are the parameters:

26-3-3
Find volume of the equation
y = 2 - x
rotated about X axis using disk method, boundaries are y = 0 and x = 0
Formula: V = pi (int) y^2 dx

Attempt:
since y = 2 - (2) = 0, 2 is an x-coordinate boundary. So the upper and lower limits of this integral are 2 and 0. X axis intercepts. Correct?

from there:
= pi (int) (2-x)^2
(2-x)^2 = 4 - 3x

= pi (int) 4 - 3x dx
= pi/3 (int) 4 - x dx

Integration:

= pi/3 4x - 1/2(x^2)

Plugging in 2 = 6.3776

However, the answer is 8.776. Any idea where I went wrong? thanks.
• Jul 25th 2013, 03:35 PM
MarkFL
Re: Integration of a volume
One problem is where you expanded the integrand...$\displaystyle (2-x)^2\ne4-3x$.
• Jul 25th 2013, 04:46 PM
HallsofIvy
Re: Integration of a volume
Do you really need Calculus to do this? y= 2- x is a straight line. When x= 0, y= 2. When y= 0, x= 2. Rotating around the x-axis gives a cone with height 2 and radius 2. The volume of a cone of height h and radius r is given by $\displaystyle V= (1/3)\pi r^2 h$.
• Jul 27th 2013, 06:20 PM
togo
Re: Integration of a volume
No don't need calculus, however it is a calculus question and they will get harder.

It was my understanding that everything to the right of the integral symbol you simplify algebraically and then integrate but obviously not