I have a maths test coming up, and I just can't seem to solve a question on finding volumes of solids (via integration). Here's the question:
Find the volume of a pyramid with height h and rectangular base with dimensions b and 2b.
As straightforward as it seems, I am very confused. I would really be grateful for some help. If possible, please help me understand the solution using a simple diagram.
Thanks a lot
Best Answer
Method 1: Using Calculus
Consider the pyramid having apex point at the origin & its axis coinciding with the x-axis then at a distance $x$ from the origin, consider an elementary cuboid having small thickness $dx$ & a rectangular cross-section of width $b_x=\frac{bx}{h}$ & length $lx=\frac{2bx}{h}$
Then the volume of elementary cuboid $$dV=\text{(area of rectangular cross section)}\times {(thickness)}$$ $$dV=b_xl_xdx=\frac{bx}{h}\cdot \frac{2bx}{h}\cdot dx=\frac{2b^2}{h^2}x^2dx$$ Hence, the total volume of the pyramid $$V=\int dV=\int \frac{2b^2}{h^2}x^2dx$$ Using the proper limits of variangle $x$, we get volume of complete pyramid as follows $$V=\int_{0}^{h}\frac{2b^2}{h^2}x^2dx$$ $$=\frac{2b^2}{h^2}\int_{0}^{h}x^2dx$$ $$=\frac{2b^2}{h^2}\left[\frac{x^3}{3}\right]_{0}^{h}$$ $$=\frac{2b^2}{3h^2}\left[h^3-0\right]$$ $$=\frac{2}{3}b^2h$$
Method 2: Using Geometry
Volume of the right pyramid with rectangular base $$=\frac{1}{3}(\text{area of rectangular base})\times(\text{vertical height})$$ $$=\frac{1}{3}(b\cdot2b)\times (h)$$ $$=\frac{2}{3}b^2h$$