Would that give me the area for S? The formula that I took a picture of.
Im doing (b)
I dont think so.
This gives you the right soln
$\displaystyle \int_3^6 y dx - \int_3^4 f(x) dx $
Basically get the area under the curve of the line from where the line is tangent to f(x)(x=3) to where the line intersects the x-axis and subtract that from, the point where the line is tangent to f(x) (x=3) to where f(x) intersects the x-axis (x=4)
The tangent point is x=3, f(x) crosses the x-axis at x=4, and the line crosses the x-axis at x=6. So if you divide the area S into two regions with a vertical line at x=4, you have $\displaystyle S=\int_3^4 (18-3x)-(4x^2-x^3)\,dx + \int_4^6 18-3x\,dx$.
And I think that gives the same answer as jakncoke.
I think there's an extra 2 in ibdutt's answer to part (c) - it should be $\displaystyle \int_0^4 \pi(4x^2-x^3)^2\,dx$.
- Hollywood