Answer:
The volume of this pyramid is 16 cm³.
Step-by-step explanation:
The volume [tex]V[/tex] of a solid pyramid can be given as:
[tex]\displaystyle V = \frac{1}{3} \cdot b \cdot h[/tex],
where
- [tex]b[/tex] is the area of the base of the pyramid, and
- [tex]h[/tex] is the height of the pyramid.
Here's how to solve this problem with calculus without using the previous formula.
Imaging cutting the square-base pyramid in half, horizontally. Each horizontal cross-section will be a square. The lengths of these squares' sides range from 0 cm to 3 cm. This length will be also be proportional to the vertical distance from the vertice of the pyramid.
Refer to the sketch attached. Let the vertical distance from the vertice be [tex]x[/tex] cm.
- At the vertice of this pyramid, [tex]x = 0[/tex] and the length of a side of the square is also [tex]0[/tex].
- At the base of this pyramid, [tex]x = 3[/tex] and the length of a side of the square is [tex]4[/tex] cm.
As a result, the length of a side of the square will be
[tex]\displaystyle \frac{x}{3}\times 4 = \frac{4}{3}x[/tex].
The area of the square will be
[tex]\displaystyle \left(\frac{4}{3}x\right)^{2} = \frac{16}{9}x^{2}[/tex].
Integrate the area of the horizontal cross-section with respect to [tex]x[/tex]
- from the top of the pyramid, where [tex]x = 0[/tex],
- to the base, where [tex]x = 3[/tex].
[tex]\displaystyle \begin{aligned}\int_{0}^{3}{\frac{16}{9}x^{2}\cdot dx} &= \frac{16}{9}\int_{0}^{3}{x^{2}\cdot dx}\\ &= \frac{16}{9}\cdot \left(\frac{1}{3}\int_{0}^{3}{3x^{2}\cdot dx}\right) & \text{Set up the integrand for power rule}\\ &= \left.\frac{16}{9}\times \frac{1}{3}\cdot x^{3}\right|^{3}_{0}\\ &= \frac{16}{27}\times 3^{3} \\ &= 16\end{aligned}[/tex].
In other words, the volume of this pyramid is 16 cubic centimeters.
