Use calculus to find the volume of a pyramid with height h and rectangular base with dimensions b and 2b.
I don't know how to start this question. I've tried finding the integral from 0 to h of (b)(2b)db, but that doesn't seem to work.
Use calculus to find the volume of a pyramid with height h and rectangular base with dimensions b and 2b.
I don't know how to start this question. I've tried finding the integral from 0 to h of (b)(2b)db, but that doesn't seem to work.
Take a horizontal cross-section of the pyramid. Let $\displaystyle s$ be the smaller side of the cross-section and let $\displaystyle x$ be the height from the cross-section to the base of the pyramid. Due to similarity, the longer side of the cross-section must have length $\displaystyle 2s$ and the area will be $\displaystyle 2s^2$. Therefore, the volume of the pyramid will be
$\displaystyle V=\int_{0}^{h}2s^2dx$.
To find $\displaystyle s$ in terms of $\displaystyle x$, use the similarity of triangles (vertically 'slice' the pyramid in half and examine the cross-section).
Place this imaginary pyramid on a co-ordinate system, with the apex of the pyramid at the origin. The sloping side of the pyramid forms, well a slope, with points (0,0) and (h, b) (if we use 2b, then half of its height above the x-axis will be b). Therefore an equation of the line for this pyramid would be $\displaystyle y=\frac{b}{h}x$, and the area of a cross section of this pyramid would be $\displaystyle A(x)=\left(\frac{2b}{h}x\right)\left(b\right)\Righ tarrow A(x)=\left(\frac{2b^{2}x}{h}\right)$
It is then simply a matter of integrating the area function using 0 and h as limits of integration to find the total volume of the pyramid.