Problem:
Find the volume of the solid whose base is the region in the xy-plane that is bounded by the parabola
$y = 4 - x^2$ and the line $y = 3x$, while the top of the solid is bounded by the plane $z = x + 4$.
Answer:
Let $V$ be the volume we seek. Observe that at $x = 0$ we have $(4-x^2) > 3x$. To find the limits of integration
for $x$, we setup the following equation.
\begin{align*}
4 - x^2 &= 3x \\
-x^2 - 3x + 4 &= 0 \\
x^2 + 3x - 4 &= 0 \\
x &= \dfrac{ -3 \pm \sqrt{ 9 - 4(1)(-3) } }{2(1)} \\
x &= \dfrac{ -3 \pm \sqrt{ 25 } }{2} \\
x &= 1 \text{ or } x = -4 \\
V &= \int_{-4}^1 \int_{3x}^{4-x^2} x + 4 \,\, dy \,\, dx \\
V &= \int_{-4}^1 (x + 4)y \Big|_{3x}^{4-x^2} \,\, dx \\
V &= \int_{-4}^1 (x + 4)(4-x^2-3x) \,\, dx \\
V &= \int_{-4}^1 4x - x^3 - 3x^2 + 16 - 4x^2 - 12x \,\, dx
\end{align*}
Using an online integral calculator, I find that: $A = -\dfrac{2345}{12}$.
However, the book's answer is $\dfrac{625}{12}$. Where did I go wrong?
The volume is the integral
$$ V=\iiint_V dV $$
Your question is over the area bounded by the curves $y=3x$ and $y=4-x^2$ and between the planes $z=0$ and $z=x+4$. So the integral is written as:
$$ V= \int_{x=-4}^1 \int_{y=3x}^{4-x^2} \int_{z=0}^{x+4} dz\ dy\ dx $$
You correctly work the integral and get to the point of
$$ V=\int_{-4}^1 16-8x-7x^2-x^3 dx = [16x-4x^2-7/3x^3-1/4x^4]_{-4}^1=625/12 $$
Which is the book's answer.