Answer:
[tex]V=\frac{512\pi }{15}[/tex]
Step-by-step explanation:
First thing I did was graph this on my calculator to find the upper and lower bounds. They were x = 5, 9. Then I applied the disk method of integration. From this I know know I use y = x equations, x intervals, I need a value for R, and then I need a value for r. R represents the height of the representative rectangle which is perpendicular to the axis of rotation, so the height of that rectangle starts at y = 0 and meets the curve. So
[tex]R=-x^2+14x-45[/tex]
Since there is no "hole" or space between the graph and the axis of rotation, r=0. Putting that into the volume formula:
[tex]V=\pi \int\limits^9_5 {[-x^2+14x-45]^2-[0]^2} \, dx[/tex]
To simplify I have to multiply that polynomial by itself. Doing that gives us:
[tex]V=\pi \int\limits^9_5 {x^4-28x^3+286^2-1260x+2025} \, dx[/tex]
Integrating gives us:
[tex]V=\pi [\frac{x^5}{5}-\frac{28x^4}{4}+\frac{286x^3}{3}-\frac{1260x^2}{2}+2025x[/tex] which we integrate from 5 to 9
Doing a bit of simplification on that:
[tex]V=\pi [\frac{x^5}{5}-7x^4+\frac{286x^3}{3}-630x^2+2025x][/tex] from 5 to 9
Applying the First Fundamental Theorem of Calculus we get that when we sub in a 9 for x then a 5 for x:
[tex]V=\pi(\frac{12879}{5}-\frac{7625}{3})[/tex]
which subtracts to
[tex]V=\frac{512\pi }{15}[/tex]
