Aimee H. answered 03/21/13
This was a little easier than I thought, but may or may not be useful since your chapter is to specifically use integration.
The equation for the cylinder, x^2+y^2=4 is in the format a^2 + b^2 = r^2, therefore, r=2. As for height, I'm not sure why you're asking for height. To solve this, you can either do a double integration or a way that doesn't involve integration at all. As far as the double integration, I only got to A(s)=SSD sqrt(1+(-3/2)^2+(-1/2)^2) dA , -3/2 being fx and -1/2 being fy of the plane if you solve for z. (SS is my poor shorthand for the two integrals). But I don't actually have a cal3 book, so I didn't get very far, and you'd have to integrate the surface area.
So instead, I used the method without integration, in which I took the normal of the plane, n=(3,1,2) and the normal of the xy axis, (0,0,1), and found the angle between them.
cos(a)=(3(0)+1(0)+1(2))/(sqrt(3^2+1^2+2^2)*sqrt(0^2+0^2+1^2)) from the cos(a)= n1*n2 / (|n1||n2|)
Since the cylinder is circular, we can use the area of a circle on the xy plane by projecting the area and set it up like this:
A= Axy / cos(a) = Axy sqrt(14) / 2
Since the area of a circle is pi*r^2, we can fill that in for Axy. Remember, our r=2.
A= pi*r^2 *sqrt(14)/2 = 2*pi*sqrt(14)
Thank you, guys. I'll search for the right way to do it.