Question

The region bounded by the given curves is rotated about the specified axis. Find the volume V of the resulting solid by any method. y = −x2 + 14x − 45, y = 0; about the x-axis

Answer 1

Answer:

$V=\frac{512\pi }{15}$

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

$R=-x^2+14x-45$

Since there is no "hole" or space between the graph and the axis of rotation, r=0.  Putting that into the volume formula:

$V=\pi \int\limits^9_5 {[-x^2+14x-45]^2-[0]^2} \, dx$

To simplify I have to multiply that polynomial by itself.  Doing that gives us:

$V=\pi \int\limits^9_5 {x^4-28x^3+286^2-1260x+2025} \, dx$

Integrating gives us:

$V=\pi [\frac{x^5}{5}-\frac{28x^4}{4}+\frac{286x^3}{3}-\frac{1260x^2}{2}+2025x$ which we integrate from 5 to 9

Doing a bit of simplification on that:

$V=\pi [\frac{x^5}{5}-7x^4+\frac{286x^3}{3}-630x^2+2025x]$ 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:

$V=\pi(\frac{12879}{5}-\frac{7625}{3})$

which subtracts to

$V=\frac{512\pi }{15}$