Find the surface of the paraboloid y=x²+z² that lies inside the cylinder x²+z²=9.

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 simple substitute

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

Overt the surface x² + z² = 9.

Where θ varies from 0 to 2π and

r varies from 0 to 3.

S = \( \int_{\theta =0}^{1}\;\int_{r=0}^{3}\sqrt{1 + 4r^{2}} \)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 = 3 then t = 37

S = \( \int_{\theta =0}^{2\pi }\;\int_{t=1}^{37}\)t × tdt/4 dθ

= 1/4 \( \int_{\theta =0}^{2\pi }\;\int_{t=1}^{37}\) t²dt dθ

=1/4 \( \int_{\theta =0}^{2\pi }\)t³ /3]\(_{t=1}^{37}\) dθ

= (1/4)(50653/3 - 1/3)\( \int_{\theta =0}^{2\pi }\) dθ

= (1/4)(50652/3)

= (1/4)(16884)(2π)

S = 8442π

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Find the surface of the paraboloid y=x²+z² that lies inside the cylinder x²+z²=9.

Summary:

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

x²+z² = 1 is 8442π