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# Find the area of the surface. The part of the plane x+2y+3z=1 that lies inside the cylinder x2 + y2=3.

**Solution:**

Given plane x+2y+3z=1 and cylinder x^{2} + y^{2 }= 3

Let us assume P is the projection on the plane x+2y+3z=1

The surface area of an equation is given by

∫∫ₚ √{1+(∂z/∂x)² + (∂z/∂y)²} dA

The plane is inside the cylinder.

By using cylindrical coordinates:

x = r cos θ, y = r sin θ

x^{2} + y^{2} = (rcos θ)^{2} +(rsin θ)^{2}

= r^{2} (cos^{2} θ + sin^{2 }θ)

= r^{2} . 1 = r^{2}

r^{2} = 3 ⇒ r = √3

Thus P = {(r,θ)/ 0 ≤ r ≤ √3, 0 ≤ θ ≤ 2π}

x +2y + 3z =1

⇒ z = 1/3(1 - x - 2y)

∂z/∂x = -1/3 ; ∂z/∂y = -2/3

Surface Area = ∫∫ₚ √{1+(∂z/∂x)² + (∂z/∂y)²} dA

= ∫∫ₚ √{1+ (-⅓) ² + (-⅔) ²} dA

= ∫∫ₚ √(9+1+4)/ √9 dA

= √14 /3 \(\int\limits_0^{2π}\int\limits_0^{√3}\) dA is area of region P

= √14 /3 \(\int\limits_0^{2π} \int\limits_0^{√3}\) r. dr. dθ

= √14 /3 \(\int\limits_0^{2π}\) [r^{2 }/2]\(^{√3}_0\) .dθ

= √14 /3 \(\int\limits_0^{2π}\) 3/2 .dθ

= √14/2 [θ]\(^{2π}_0\)

=√14/2 [2π - 0]

= π√14

Area of the plane that lies inside the given cylinder : (√14 /3) × 3π = π√14 sq. units

## Find the area of the surface. The part of the plane x+2y+3z=1 that lies inside the cylinder x^{2} + y^{2}=3.

**Summary: **

The area of the part of the plane x+2y+3z=1 that lies inside the cylinder x^{2} + y^{2}=3 is π√14 sq. units.

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