# Find the surface of the paraboloid y=x^{2}+z^{2} that lies inside the cylinder x^{2}+z^{2}=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 = x^{2} + z^{2}

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 x^{2} + z^{2} = 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 t^{2 }= 1 + 4r^{2} 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 \( \int_{\theta =0}^{2\pi }\;\int_{t=1}^{37}\) t^{2}dt dθ

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

## Find the surface of the paraboloid y=x^{2}+z^{2} that lies inside the cylinder x^{2}+z^{2}=9.

**Summary: **

The Surface Integral.S is the part of the paraboloid y = x^{2}+z^{2} that lies inside the cylinder

x^{2}+z^{2} = 1 is 8442π

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