<s>[INST] Finding the volume when a parabola is rotated about the line $y = 4$. Problem:
Find the volume generated by revolving the region bounded below by the
parabola $y = 3x^2 + 1$ and above by the line $y = 4$ about the line $y = 4$.
Answer:
Let $V$ be the volume we are trying to find. We want to find the points where the two curves intersect. Hence,
set up the following equation:
$$ 3x^2 + 1 = 4$$.
From this equation, we find two solutions: $ x = \pm 1 $
\begin{align*}
du &= -dx \\
V &= \pi \int_{-1}^{1} (4 - (3x^2 + 1))^2 \,\,\, dx = \pi \int_{-1}^{1} (3 - 3x^2)^2 \,\,\, dx \\
V &= 9 \pi \int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx \\
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= \int_{-1}^{1} x^4 - 2x^2 + 1 \,\,\, dx \\
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= \frac{x^5}{5} - \frac{2x^3}{3} + x \Big|_{-1}^{1} \\
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\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &=
 \frac{^5}{5} - \frac{2(1)^3}{3} + 1 - \left( \frac{(-1)^5}{5} - \frac{2(-1)^3}{3} - 1 \right) \\
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\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= \frac{1}{5} - \frac{2}{3} + 1 - \left( -\frac{1}{5} + \frac{2}{3} - 1 \right) \\
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= \frac{1}{5} - \frac{2}{3} + 1 + \frac{1}{5} - \frac{2}{3} + 1 \\
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= 2 - \frac{4}{3} + \frac{2}{5}  = 2 - \frac{20}{15} + \frac{6}{15} \\
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= \frac{16}{15} \\
V &= 9 \pi \left( \frac{16}{15} \right) \\
V &= \frac{ 48 \pi }{5}
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\end{align*}
The book's answer is:
$$ \frac{144 \pi}{15} $$
I claim my answer is right. That is, the book failed to simplify its answer. Am I missing something? Please comment.
 [/INST] $$\frac{144 \pi}{15}$$
dividing the numerator and denominator by three:
$$\frac{48 \pi}{5}$$
In the future, if you don't feel like simplifying, you can just divide the two answers and see if it lines up
$$\frac{144 \pi}{15} \approx 30.1593$$
$$\frac{48 \pi}{5} \approx 30.1593$$
So, yes, you're right.
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