Evaluate the Surface Integral.S is the part of the paraboloid y = x2 + z2 that lies inside the cylinder x2 + z2 = 1.

Solution:

The area of surface y , above the region R of the cartesian form is given by

S = \( \iint_{R}^{}\sqrt{1+(dy/dx)^{2}+(dy/dz)^{2}} \) dA

Given paraboloid y = x2 + z2

dy/dx= 2x, dy/dz = 2z

S = \( \iint_{R}^{}\sqrt{1+(dy/dx)^{2}+(dy/dz)^{2}} \) dA

S = \( \iint_{R}^{}\sqrt{1+(2x)^{2}+(2z)^{2}} \) dxdz

S = \( \iint_{R}^{}\sqrt{1+4(x)^{2}+4(z)^{2}} \) dxdz

S = \( \iint_{R}^{}\sqrt{1+4(x^{2}+z^{2}}) \) dxdz

In order to make the integration simply substitute

X = rcosθ and z = r sinθ and dxdz = r dr dθ

Over the surface x2 + z2 = 1.

Where θ varies from 0 to 2π and

r varies from 0 to 1.

S = \( \int_{\theta =0}^{1}\;\int_{r=0}^{1}\sqrt{1 + 4r^{2}} \)r dr dθ

If t2= 1 + 4r2 then 8r dr = 2tdt

r dr = t dt / 4 and if r = 0 then t =1 and if r = 1 then t = 5

S = \( \int_{\theta =0}^{2\pi }\;\int_{t=1}^{5}\)t × tdt/4 dθ

= 1/4 \( \int_{\theta =0}^{2\pi }\;\int_{t=1}^{5}\) t2dt dθ

=1/4 \( \int_{\theta =0}^{2\pi }\)[t3 /3]\(_{1}^{5}\)dθ

= (1/4) (125/3 - 1/3)\( \int_{\theta =0}^{2\pi}\) dθ

= (1/4)(124/3)

= (1/4)(8)(2π)

S = 4π

Evaluate the Surface Integral.S is the part of the paraboloid y = x2 + z2 that lies inside the cylinder x2 + z2 = 1.

Summary:

The Surface Integral.S is the part of the paraboloid y = x2 + z2 that lies inside the cylinder x2 + z2 = 1 is 4π

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