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.