Question

(Calculus) 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 = −x^2 + 14x − 45, y = 0; about the y−axis

Answer 1

\[V = \int\limits_{5}^{9}2\pi x ( -x^2+14x-45)dx = 2\pi \int\limits_{5}^{9} x ( -x^2+14x-45)dx\]
\[= 2\pi[\frac{ -1 }{ 4 }x^4+\frac{ 14 }{ 3 }x^3-\frac{ 45 }{ 2 }x^2]_{5}^{9}\]

Is this correct so far?

Answer 2

\[=2\pi[(\frac{ -6561 }{ 4 }+3402-\frac{ 3645 }{ 2 })-(-\frac{ 625 }{ 4 }+\frac{ 1750 }{ 3 }-\frac{ 1125 }{ 2 })\]
\[=2\pi[(\frac{ -19683+40824-21870)-(-1875+7000-6750 )}{ 12 }]\]
\[=2\pi [ ( \frac{ -729 }{ 12 })-(-\frac{ 1625 }{ 12 }) = 2\pi [ \frac{ 896 }{ 6 }] = \frac{ 896 }{ 3 }\pi\]

My answer is wrong, and I don't know where I goofed at.

Answer 3

I believe you've found surface area rather than volume