# Simple integration problem.

• Mar 26th 2009, 07:35 AM
craigmain
Simple integration problem.
What is the cost of a swimming pool cover. the pool is bounded by $\displaystyle y=x^2$ and $\displaystyle y=2$ and it costs $2 / foot for the cover. One unit of the x and y axes are 50 feet. I have tried the following: Area under the graph$\displaystyle y=x^2\displaystyle \int\limits_{ - \sqrt 2 }^{\sqrt 2 } {{x^2}dx}$Which is$\displaystyle \left. {\frac{{{x^3}}}{3}} \right|_{ - \sqrt 2 }^{\sqrt 2 }$Which equals$\displaystyle \frac{4\sqrt 2}{3}$. Area of cover is rectangle less area under the graph.$\displaystyle 4\sqrt{2} -\frac{4\sqrt 2}{3}$That multiplied by 50, and then by 2$\displaystyle \frac{8\sqrt 2}{3}(50)(2)$Equals$\displaystyle \approx{377.123}$My textbook has$18,856. I am not even in the same ball park.

Thanks
Regards
Craig.
• Mar 26th 2009, 08:47 AM
chisigma
Simply you forgot that tha area is in foot square... that means you have to multiply by 50 again the result and you obtain 50 x 377.123 = 18,586... just as in your textbook (Itwasntme) ...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$