1. ## volume of solid

What i the volume of the solid
y=3x-x^2, x=3, y=0, revolved about the y-axis

2. What have you tried? How far have you gotten on this exercise?

3. do i use the formula
V=(2pie)interval from a to b ((f(x)-(g(x)))dx

4. If thats the right formula i get the answer to be 18pie/2

5. Originally Posted by chevy900ss
What i the volume of the solid
y=3x-x^2, x=3, y=0, revolved about the y-axis
Shell method

$\displaystyle {V_y} = 2\pi \int\limits_a^b {xydx} = 2\pi \int\limits_0^3 {x\left( {3x - {x^2}} \right)dx} = 2\pi \int\limits_0^3 {\left( {3{x^2} - {x^3}}\right)dx} =$

$\displaystyle = \left. {2\pi \left( {{x^3} - \frac{{{x^4}}}{4}} \right)} \right|_0^3 = 2\pi \left( {27 - \frac{{81}}{4}} \right) = 2\pi \cdot \frac{{27}} {4} = \frac{{27}}{2}\pi .$

6. Ok i forgot that x in the beginning equation.
So, with the problem y=(x^3)-(5x^2)+(6x), x=2, y=0
the answer for this one would be 24pie/5

7. Originally Posted by chevy900ss
Ok i forgot that x in the beginning equation.
So, with the problem y=(x^3)-(5x^2)+(6x), x=2, y=0
the answer for this one would be 24pie/5
Yes!

$\displaystyle {V_y} = 2\pi \int\limits_0^2 {x\left( {{x^3} - 5{x^2} + 6x} \right)dx} = \ldots = \frac{{24}}{5}\pi .$

See picture