Science:Math Exam Resources/Courses/MATH101/April 2005/Question 05 (a)/Solution 1

Consider the following diagram:

In the picture, we take the perpendicular bisector of the top base and draw the line. This splits our equilateral triangle into two equal parts. The base is ${\sqrt {3}}$ and the height is 3 and the other side length is $2{\sqrt {3}}$. Choose a sample point say $\displaystyle x_{i}^{*}$. Let's call the length of the small triangle point downward l. By similar triangles, we have

$\displaystyle {\frac {\sqrt {3}}{3}}={\frac {l/2}{3-x_{i}^{*}}}$

and hence

$\displaystyle l={\frac {2(3-x_{i}^{*}){\sqrt {3}}}{3}}$

The volume of the thin prism at this height is

$\displaystyle V_{i}=l(12)\Delta x={\frac {2(3-x_{i}^{*}){\sqrt {3}}}{3}}(12\Delta x)=8(3-x_{i}^{*}){\sqrt {3}}\Delta x$

Next, the force acting on this prism is

$\displaystyle F_{i}=V_{i}(density)=(62)8(3-x_{i}^{*}){\sqrt {3}}\Delta x$

Lastly, the work done on this piece, noting that the displacement is 1 added to the distance to the bottom of the triangle (which is $\displaystyle (3-x_{i}^{*})$)

$\displaystyle W_{i}=F_{i}(displacement)=(62)8(3-x_{i}^{*}){\sqrt {3}}\Delta x(4-x_{i}^{*})$

Summing over all pieces, we have

$\displaystyle W=\lim _{n\to \infty }\sum _{i=1}^{n}W_{i}=\lim _{n\to \infty }\sum _{i=1}^{n}(62)8(3-x_{i}^{*}){\sqrt {3}}\Delta x(4-x_{i}^{*})$

and this last value is an integral given by

${\begin{aligned}W&=\int _{0}^{3}(62)8(3-x){\sqrt {3}}(4-x)\,dx\\&=62(8){\sqrt {3}}\int _{0}^{3}(12-7x+x^{2})\,dx\\&=62(8){\sqrt {3}}(12x-7x^{2}/2+x^{3}/3){\bigg |}_{0}^{3}\\&=62(8){\sqrt {3}}(12(3)-7(3)^{2}/2+(3)^{3}/3-0)\\&=62(8){\sqrt {3}}(36-63/2+9)\\&=62(4)(27){\sqrt {3}}\end{aligned}}$