Science:Math Exam Resources/Courses/MATH101/April 2006/Question 05/Solution 1

Start by placing a coordinate system on your picture with 0 on top. We break the problem up into two parts. First, we take care of the square.

Choose a sample point $\displaystyle x_{i}^{*}$ somewhere between 0 and 50 on your picture. Draw the horizontal line across at this point. A picture is included below.

The area of this horizontal strip is given by

$\displaystyle A_{i}=50\Delta x$.

The pressure on this strip is given by

$\displaystyle P_{i}=\rho gx_{i}^{*}$.

where $\displaystyle \rho =1000$ is the density of water and $\displaystyle g=9.8$ is the acceleration due to gravity. Thus, the force on this strip is

$\displaystyle F_{i}=P_{i}A_{i}=50\rho gx_{i}^{*}\Delta x$.

The total force is

$\displaystyle F_{sq}=\lim _{n\to \infty }\sum _{i=1}^{n}F_{i}=\lim _{n\to \infty }\sum _{i=1}^{n}50\rho gx_{i}^{*}\Delta x$.

This can be expressed as the following integral

$\displaystyle F_{sq}=\int _{0}^{50}50\rho gx\,dx$

Now we compute the force on the semicircle and add it to this force to get the total force. To make the notation easier, I will reuse some of the letters from above.

Choose a sample point $\displaystyle x_{i}^{*}$ somewhere between 50 and 75 on your picture. Draw the horizontal line across at this point. A picture is included below.

To compute the length of the horizontal line, draw the radius as shown in the picture. Then the Pythagorean theorem tells us that the length across is double $\displaystyle {\sqrt {25^{2}-(x_{i}^{*}-50)^{2}}}$. Thus, the area of this strip is given by

$\displaystyle A_{i}=2{\sqrt {25^{2}-(x_{i}^{*}-50)^{2}}}\Delta x$.

The pressure on this strip is given by

$\displaystyle P_{i}=\rho gx_{i}^{*}$.

Thus, the force on this strip is

$\displaystyle F_{i}=P_{i}A_{i}=2{\sqrt {25^{2}-(x_{i}^{*}-50)^{2}}}\rho gx_{i}^{*}\Delta x$.

The total force is

$\displaystyle F_{circ}=\lim _{n\to \infty }\sum _{i=1}^{n}F_{i}=\lim _{n\to \infty }\sum _{i=1}^{n}2{\sqrt {25^{2}-(x_{i}^{*}-50)^{2}}}\rho gx_{i}^{*}\Delta x$.

This can be expressed as the following integral

$\displaystyle F_{circ}=\int _{50}^{75}2{\sqrt {25^{2}-(x-50)^{2}}}\rho gx\,dx$

Adding the two integrals together, we get

${\begin{aligned}F_{tot}&=F_{sq}+F_{circ}\\&=\int _{0}^{50}50\rho gx+\int _{50}^{75}2{\sqrt {25^{2}-(x-50)^{2}}}\rho gx\,dx\\&={\frac {50\rho gx^{2}}{2}}{\bigg |}_{0}^{50}+\int _{0}^{25}2{\sqrt {25^{2}-x^{2}}}\rho g(x+50)\,dx\\&=25\rho g(50)^{2}+\int _{0}^{25}2{\sqrt {25^{2}-x^{2}}}\rho gx\,dx+\int _{0}^{25}100{\sqrt {25^{2}-x^{2}}}\rho g\,dx\\\end{aligned}}$

The last integral can be recognized as the [scaled] area of a quarter circle of a circle with radius 25. The first integral we can let $\displaystyle u=25^{2}-x^{2}$. When x = 0, then u= 625 and when x = 25 then u=0 and $\displaystyle du=-2xdx$. Hence,

${\begin{aligned}F_{tot}&=25\rho g(50)^{2}-\int _{625}^{0}{\sqrt {u}}\rho g\,du+100\rho g{\frac {\pi (25)^{2}}{4}}\\&=25\rho g(50)^{2}+\int _{0}^{625}{\sqrt {u}}\rho g\,du+25\rho g\pi (25)^{2}\\&=25\rho g(50)^{2}+{\frac {2u^{3/2}\rho g}{3}}{\bigg |}_{0}^{625}+\rho g\pi (25)^{3}\\&=25\rho g(50)^{2}+{\frac {2(25)^{3}\rho g}{3}}+\rho g\pi (25)^{3}\\\end{aligned}}$

which is our final answer.