← 2016 Paper 1

UPSC 2016 Maths Optional Paper 1 Q3c — Step-by-Step Solution

15 marks · Section A

Areas, surface areas, volumes via integration · Calculus · asked 4× in 13 yrs · Read the full method →

Question

Find the surface area of the plane x+2y+2z=12x+2y+2z=12 cut off by x=0, y=0x=0,\ y=0 and x2+y2=16x^2+y^2=16.

Technique

Surface area of a graph z=z(x,y)z=z(x,y) over region RR is ∬R1+zx2+zy2 dA\iint_R\sqrt{1+z_x^2+z_y^2}\,dA; for a plane the radicand is constant, so the area is (constant factor) ×\times (projected area).

Solution

Step 1 — Express the surface as z=z(x,y)z=z(x,y) and find the area element

From x+2y+2z=12x+2y+2z=12, z=12−x−2y2z=\dfrac{12-x-2y}{2}, so

zx=−12,zy=−1.z_x=-\tfrac12,\qquad z_y=-1.

The surface-area element is

dS=1+zx2+zy2 dA=1+14+1 dA=94 dA=32 dA.dS=\sqrt{1+z_x^2+z_y^2}\,dA=\sqrt{1+\tfrac14+1}\,dA=\sqrt{\tfrac94}\,dA=\tfrac32\,dA.

Step 2 — Identify the projected region RR in the xyxy-plane

The boundaries x=0x=0, y=0y=0, and the cylinder x2+y2=16x^2+y^2=16 project to the quarter disc of radius 44 in the first quadrant:

R={(x,y):x≥0, y≥0, x2+y2≤16}.R=\{(x,y):x\ge0,\ y\ge0,\ x^2+y^2\le16\}.

Its area is 14π(4)2=4π\dfrac14\pi(4)^2=4\pi.

Step 3 — Integrate

Area=∬R32 dA=32⋅Area⁡(R)=32⋅4π.\text{Area}=\iint_R \tfrac32\,dA=\tfrac32\cdot\operatorname{Area}(R)=\tfrac32\cdot4\pi.

Answer

  Surface area=6π  (≈18.85).  \boxed{\;\text{Surface area}=6\pi\;(\approx18.85).\;}
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