Re: Integration of a volume

One problem is where you expanded the integrand...$\displaystyle (2-x)^2\ne4-3x$.

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 $\displaystyle V= (1/3)\pi r^2 h$.

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