Problem:
The vertex of a pyramid lies at the origin, and the base is perpendicular to the x-axis at $x = 4$. The cross sections of the pyramid perpendicular to the x-axis are squares whose diagonals run from the curve $y = -5x^2$ to the curve $y = 5x^2$. Find the volume of the pyramid.
Answer:
The first step is to graph the two curves. A graph can be found at the following URL:
https://www.wolframalpha.com/input/?i=plot+5x%5E2+%2C+-5x%5E2
Since the cross sections of the pyramid are perpendicular to the x-axis we will integrate with respect to x. Since each cross section is a square and $y$ goes from $-5x^2$ to $5x^2$, the the length of one side of the square will be $5x^2 – (-5x^2)$ or
$10x^2$. Let $V$ be the volume we are looking for.
\begin{align*}
du &= -dx \\
V &= \int_0^4 (10x^2)^2 \,\,\, dx = \int_0^4 100x^4 \,\,\, dx \\
V &= \frac{100x^5}{5} \Big|_0^4 = 20x^5 \Big|_0^4 \\
V &= 20(4^5) = 20(1024) = 20480
%
\end{align*}
The book's answer is:
$$ 10,240 $$
Where did I go wrong?
Best Answer
Your method is correct but you lost a $1/2$ factor. The value of $10x^2$ is the diagonal, not the side of the square. So the integrand becomes $\frac 12 (10x^2)^2$.