Surface area formula
A(S)=∬D[fx(x,y)]2+[fy(x,y)]2+1dAA(S)=\iint^{\ }_{D}{\sqrt{\left\lbrack f_{x}(x,\ y)\right\rbrack ^{2}+\left\lbrack f_{y}(x,\ y)\right\rbrack ^{2}+1}\ dA} =∬D1+(∂z∂x)2+(dzdy)2dA=\iint^{\ }_{D}{\sqrt{{1+\left(\frac{\partial z}{\partial x}\right)}^{2}+\left(\frac{dz}{dy}\right)^{2}}\ dA}
Find the area of the surface.
The part of the surface 2y+4z−x2=52y+4z-x^{2}=5 that lies above the triangle with vertices (0,0)(0,\ 0), (2,0)(2,\ 0), and (2,4)(2,\ 4).
The part of the sphere x2+y2+z2=4zx^{2}+y^{2}+z^{2}=4z that lies inside the paraboloid z=x2+y2z=x^{2}+y^{2}.