# Average value of f(x,y) = x + y + 2 on the curve C_1

• Feb 17th 2014, 07:44 PM
over9000
Average value of f(x,y) = x + y + 2 on the curve C_1
Consider \$\displaystyle C_1\$ as the curve given by the semicircle of radius two in the upper half plane. Give \$\displaystyle C_1\$ the standard orientation (counter clockwise). What is the average value of \$\displaystyle f(x,y)= x + y + 2\$ on \$\displaystyle C_1\$.

I'm not even sure where to start. Could I get some help, please?
• Feb 17th 2014, 07:48 PM
over9000
Re: Average value of f(x,y) = x + y + 2 on the curve C_1
why is none of my tex showing?
• Feb 17th 2014, 08:50 PM
over9000
Re: Average value of f(x,y) = x + y + 2 on the curve C_1
I think I know what to do magically now that I posted the question. I need to come up with some parametric form for C, so C is x(t) = 2cos(t) and y(t) =2sin(t)> from t=0 to t=pi/4 Now I write everything in terms of the parametrization (that's a word spell checker!) f(x(t), y(t)) = 2cos(t)+ 2sin(t) + 2 and take the integral of that from 0 to 2pi with respect to t. Now I'm not sure what I need to do to get the average of that integral? Divide it by 2pi?
• Feb 18th 2014, 09:01 AM
hollywood
Re: Average value of f(x,y) = x + y + 2 on the curve C_1
Watch your limits of integration - since its a semicircle, t goes from 0 to pi. And yes, to get the average you divide by the length of the curve, which is 2pi.

- Hollywood
• Feb 18th 2014, 10:16 AM
over9000
Re: Average value of f(x,y) = x + y + 2 on the curve C_1
Thanks Hollywood, you're right. I confused upper half of the plane for the upper right quadrant, so all the work I did needs to be adjusted.