# Simple integration problem.

• March 26th 2009, 08:35 AM
craigmain
Simple integration problem.
What is the cost of a swimming pool cover. the pool is bounded by $y=x^2$ and $y=2$ 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 $y=x^2$ $\int\limits_{ - \sqrt 2 }^{\sqrt 2 } {{x^2}dx}$ Which is $\left. {\frac{{{x^3}}}{3}} \right|_{ - \sqrt 2 }^{\sqrt 2 }$ Which equals $\frac{4\sqrt 2}{3}$. Area of cover is rectangle less area under the graph. $4\sqrt{2} -\frac{4\sqrt 2}{3}$ That multiplied by 50, and then by 2 $\frac{8\sqrt 2}{3}(50)(2)$ Equals $\approx{377.123}$ My textbook has$18,856. I am not even in the same ball park.

Thanks
Regards
Craig.
• March 26th 2009, 09:47 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

$\chi$ $\sigma$