z x = 2 x z_x = 2x z x = 2 x
z y = 2 y z_y=2y z y = 2 y
The surface area over the region R defined by x 2 + y 2 = 9 = 3 2 x^2+y^2 = 9=3^2 x 2 + y 2 = 9 = 3 2 is
S = ∫ ∫ R ( z x ) 2 + ( z y ) 2 + 1 d x d y S = \int \int_R \sqrt{(z_x)^2+(z_y)^2+1}dxdy S = ∫ ∫ R ( z x ) 2 + ( z y ) 2 + 1 d x d y
= ∫ ∫ R 4 x 2 + 4 y 2 + 1 d x d y \int \int_R \sqrt{4x^2+4y^2+1}dxdy ∫ ∫ R 4 x 2 + 4 y 2 + 1 d x d y
Then to polar coordinates
S = ∫ 0 2 π ∫ 0 3 ( 4 r 2 + 1 ) 1 / 2 r d r d θ S = \int_{0}^{2\pi} \int_0^3(4r^2+1)^{1/2}rdrd\theta S = ∫ 0 2 π ∫ 0 3 ( 4 r 2 + 1 ) 1/2 r d r d θ
= 1 12 ∫ 0 2 π ( 4 r 2 + 1 ) 3 / 2 ∣ 0 3 d θ = \dfrac{1}{12} \int_{0}^{2\pi} (4r^2+1)^{3/2}|_0^3d\theta = 12 1 ∫ 0 2 π ( 4 r 2 + 1 ) 3/2 ∣ 0 3 d θ
= 37 37 − 1 12 ∫ 0 2 π d θ = \dfrac{37\sqrt{37} -1}{12} \int_0^{2\pi} d\theta = 12 37 37 − 1 ∫ 0 2 π d θ
= 37 37 − 1 12 2 π =\dfrac{37\sqrt{37} -1}{12} 2\pi = 12 37 37 − 1 2 π
= π 37 37 − 1 6 = \pi \dfrac{37\sqrt{37}-1}{6} = π 6 37 37 − 1
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