Question

Let

S={(x,y,z)βˆˆβ„3:x2+y2+z2=4,(xβˆ’1)2+y2≀1,zβ‰₯0}.

Then, the surface area of 𝑆 equals _____________ (rounded off to two decimal
places).

Answer: 4.57_⁒_⁒_⁒_⁒_⁒_⁒_⁒_⁒_⁒_⁒_⁒_⁒_

Solution :-

We are tasked with finding the surface area of the set π‘†βŠ†β„3 , which is defined by the intersection of a sphere of radius 2, a circular cross-section (cylinder), and a non-negative Z-constraint.

Summary of the Approach

Understand the Geometry:

Sphere: π‘₯2+𝑦2+𝑧2=4 (radius R=2 )
Cylinder: (π‘₯βˆ’1)2+𝑦2≀1 (radius 1, centered at (1, 0) in the x⁒y -plane)
Constraint: 𝑧β‰₯0 (upper hemisphere)

Intersection Area
Intersection Area

Surface area and projection on xy-plane.

That’s the surface area we want to calculate.

Given thatx2+y2+z2=4⟹z2=4βˆ’(x2+y2)∴z=4βˆ’(x2+y2)(sincezβ‰₯0given)

Let’s denote the surface area of S with S.A. hence

S.⁒A.=∬R1+[f⁑x(x,y)]2+[f⁑y(x,y)]2d⁒A

We have z=f⁑(x,y)=4βˆ’(x2+y2)

hence, f⁑x(x,y)=βˆ’2⁒x24βˆ’(x2+y2)=βˆ’x4βˆ’(x2+y2)

also, f⁑y(x,y)=βˆ’2⁒y24βˆ’(x2+y2)=βˆ’y4βˆ’(x2+y2)

Therefore, S.⁒A.=∬R1+(βˆ’x4βˆ’(x2+y2))2+(βˆ’y4βˆ’(x2+y2))2d⁒A=∬R1+x24βˆ’(x2+y2)+y24βˆ’(x2+y2)d⁒A=∬R1+x2+y24βˆ’(x2+y2)d⁒A=∬R24βˆ’(x2+y2)d⁒A

Now, apply change of variable in polar co-ordinate, put x=rcosΞΈ and y=rsinΞΈ.
For region of integration,

Given that(xβˆ’1)2+y2≀1x2βˆ’2⁒x+1+y2≀1x2+y2≀2⁒x(rcosΞΈ)2+(rcosΞΈ)2≀2⁒rcosΞΈr2≀2⁒rcosΞΈr≀2cosΞΈ.
Bound for r and ΞΈ
Bound for r and ΞΈ

since the entire circle r=2sinΞΈ is traced out for βˆ’Ο€2≀θ≀π2 and since for each fixed θ∈[βˆ’Ο€2,Ο€2] , r ranges from r=0 to r=2cosΞΈ . So, we get surface area

S.⁒A.=βˆ«βˆ’Ο€2Ο€2∫02cosΞΈ24βˆ’r2rd⁒rd⁒θ

Let 4βˆ’r2=t and differentiate w. r. to r , we get 2⁒rd⁒r=βˆ’d⁒t.

Now,

S.⁒A.=βˆ’βˆ«βˆ’Ο€2Ο€2∫02cosΞΈ1td⁒td⁒θ=βˆ’βˆ«βˆ’Ο€2Ο€2[tβˆ’12+1βˆ’12+1]02cosΞΈd⁒θ=βˆ’2βˆ«βˆ’Ο€2Ο€2[4βˆ’r2]02cosΞΈd⁒θ=βˆ’2βˆ«βˆ’Ο€2Ο€2(4βˆ’(2cosΞΈ)2βˆ’4βˆ’0)d⁒θ=βˆ’4βˆ«βˆ’Ο€2Ο€2(1βˆ’cos2ΞΈβˆ’1)d⁒θ=4βˆ«βˆ’Ο€2Ο€2(1βˆ’sinΞΈ)d⁒θ=4⁒[βˆ«βˆ’Ο€2Ο€21dβ’ΞΈβˆ’(βˆ«βˆ’Ο€2Ο€2sinΞΈd⁒θ)]=4⁒[Ο€βˆ’(βˆ’βˆ«βˆ’Ο€20sinΞΈd⁒θ+∫0Ο€2sinΞΈd⁒θ)]=4⁒[Ο€βˆ’(βˆ’[βˆ’cosΞΈ]βˆ’Ο€20+[βˆ’cosΞΈ]0Ο€2)]=4⁒[Ο€βˆ’(βˆ’(βˆ’cos0+cos(βˆ’Ο€2))+(βˆ’cosΟ€2+cos0))]=4⁒[Ο€βˆ’(βˆ’(βˆ’1+0)+(βˆ’0+1))]=4⁒[Ο€βˆ’(1+1)]=4⁒[Ο€βˆ’2]=4β’Ο€βˆ’8β‰ˆ4.566.

Rounded off to two decimal places is 4.57.

So, the correct answer is β†’ 4.57_⁒_⁒_⁒_⁒_⁒_⁒_⁒_⁒_⁒_⁒_⁒_⁒_