Question

How do you find the volume generated by revolving about the x-axis, the first quadrant region enclosed by the graphs of `y = 9 - x^2` and `y = 9 - 3x` between 0 to 3?

Answer 1

`approx 152.681`

Explanation:

You are trying to find the volume the 3d figure created by revolving this region around the x-axis

We are going to use the washer method.

First we need to determine the bounds

We can do this by setting both equations equal to each other:

`9-x^2=9-3x`
`0=x^2-3x`
`0=x(x-3)`
`x=0,3`

Now we can apply the washer method

`V = pi\int_0^3(f(x)^2-g(x)^2)dx`

In this formula f(x) must be greater than g(x) over the bounds of the integral. In our scenario the function that satisfies this is `y=9-x^2`.

`therefore f(x) = 9-x^2, g(x) = 9-3x`

Now we just need to plug in f(x) and g(x) into the washer method and integrate.

`V = pi\int_0^3((9-x^2)^2-(9-3x)^2)dx`
`=pi\int_0^3((81-18x^2+x^4)-(81-54x+9x^2))dx`
`=pi\int_0^3(x^4-27x^2+54x)dx`
`=pi[x^5/5-9x^3+27x^2]_0^3`
`approx 152.681`