Find the area of the paraboloid x2 + y2 = z inside the cylinder x2 + y2 = 9.
zx=2xz_x = 2xzx=2x
zy=2yz_y=2yzy=2y
The surface area over the region R defined by x2+y2=9=32x^2+y^2 = 9=3^2x2+y2=9=32 is
S=∫∫R(zx)2+(zy)2+1dxdyS = \int \int_R \sqrt{(z_x)^2+(z_y)^2+1}dxdyS=∫∫R(zx)2+(zy)2+1dxdy
= ∫∫R4x2+4y2+1dxdy\int \int_R \sqrt{4x^2+4y^2+1}dxdy∫∫R4x2+4y2+1dxdy
Then to polar coordinates
S=∫02π∫03(4r2+1)1/2rdrdθS = \int_{0}^{2\pi} \int_0^3(4r^2+1)^{1/2}rdrd\thetaS=∫02π∫03(4r2+1)1/2rdrdθ
=112∫02π(4r2+1)3/2∣03dθ= \dfrac{1}{12} \int_{0}^{2\pi} (4r^2+1)^{3/2}|_0^3d\theta=121∫02π(4r2+1)3/2∣03dθ
=3737−112∫02πdθ= \dfrac{37\sqrt{37} -1}{12} \int_0^{2\pi} d\theta=123737−1∫02πdθ
=3737−1122π=\dfrac{37\sqrt{37} -1}{12} 2\pi=123737−12π
=π3737−16= \pi \dfrac{37\sqrt{37}-1}{6}=π63737−1
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