Question

Let
$S = {(x, y, z) \in ℝ^{3} : x^{2} + y^{2} + z^{2} = 4, {(x - 1)}^{2} + y^{2} \leq 1, z \geq 0} .$
Then, the surface area of 𝑆 equals _____________ (rounded off to two decimal
places).

Answer: $\underset{_____________}{4.57}$

Solution :-

We are tasked with finding the surface area of the set $𝑆 \subseteq ℝ^{3}$ , 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: $𝑥^{2} + 𝑦^{2} + 𝑧^{2} = 4$ (radius $R = 2$ )
Cylinder: ${(𝑥 - 1)}^{2} + 𝑦^{2} \leq 1$ (radius 1, centered at (1, 0) in the $x y$ -plane)
Constraint: $𝑧 \geq 0$ (upper hemisphere)

Surface area and projection on xy-plane.

That’s the surface area we want to calculate.

\begin{array}{llllllllll} Given that x^{2} + y^{2} + z^{2} & = & 4 \\ ⟹ z^{2} & = & 4 - (x^{2} + y^{2}) \\ ∴ z & = & \sqrt{4 - (x^{2} + y^{2})} (since z \geq 0 given) \end{array}

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

S . A . = \iint_{R} \sqrt{1 + {[f_{x} (x, y)]}^{2} + {[f_{y} (x, y)]}^{2}} d A

We have $z = f (x, y) = \sqrt{4 - (x^{2} + y^{2})}$

hence, $f_{x} (x, y) = \frac{- 2 x}{2 \sqrt{4 - (x^{2} + y^{2})}} = \frac{- x}{\sqrt{4 - (x^{2} + y^{2})}}$

also, $f_{y} (x, y) = \frac{- 2 y}{2 \sqrt{4 - (x^{2} + y^{2})}} = \frac{- y}{\sqrt{4 - (x^{2} + y^{2})}}$

\begin{array}{llllllllll} Therefore, S . A . & = & \iint_{R} \sqrt{1 + {(\frac{- x}{\sqrt{4 - (x^{2} + y^{2})}})}^{2} + {(\frac{- y}{\sqrt{4 - (x^{2} + y^{2})}})}^{2}} d A \\ = & \iint_{R} \sqrt{1 + \frac{x^{2}}{4 - (x^{2} + y^{2})} + \frac{y^{2}}{4 - (x^{2} + y^{2})}} d A \\ = & \iint_{R} \sqrt{1 + \frac{x^{2} + y^{2}}{4 - (x^{2} + y^{2})}} d A \\ = & \iint_{R} \frac{2}{\sqrt{4 - (x^{2} + y^{2})}} d A \end{array}

Now, apply change of variable in polar co-ordinate, put $x = r cos θ$ and $y = r sin θ .$
For region of integration,

\begin{array}{llllllllll} Given that {(x - 1)}^{2} + y^{2} & \leq & 1 \\ x^{2} - 2 x + 1 + y^{2} & \leq & 1 \\ x^{2} + y^{2} & \leq & 2 x \\ {(r cos θ)}^{2} + {(r cos θ)}^{2} & \leq & 2 r cos θ \\ r^{2} & \leq & 2 r cos θ \\ r & \leq & 2 cos θ . \end{array}

since the entire circle $r = 2 sin θ$ is traced out for $- \frac{π}{2} \leq θ \leq \frac{π}{2}$ and since for each fixed $θ \in [- \frac{π}{2}, \frac{π}{2}]$ , $r$ ranges from $r = 0$ to $r = 2 cos θ$ . So, we get surface area

S . A . = \int_{- \frac{π}{2}}^{\frac{π}{2}} \int_{0}^{2 cos θ} \frac{2}{\sqrt{4 - r^{2}}} r d r d θ

Let $4 - r^{2} = t$ and differentiate w. r. to $r$ , we get $2 r d r = - d t .$

Now,

\begin{array}{llllllllll} S . A . & = & - \int_{- \frac{π}{2}}^{\frac{π}{2}} \int_{0}^{2 cos θ} \frac{1}{\sqrt{t}} d t d θ \\ = & - \int_{- \frac{π}{2}}^{\frac{π}{2}} {[\frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1}]}_{0}^{2 cos θ} d θ \\ = & - 2 \int_{- \frac{π}{2}}^{\frac{π}{2}} {[\sqrt{4 - r^{2}}]}_{0}^{2 cos θ} d θ \\ = & - 2 \int_{- \frac{π}{2}}^{\frac{π}{2}} (\sqrt{4 - {(2 cos θ)}^{2}} - \sqrt{4 - 0}) d θ \\ = & - 4 \int_{- \frac{π}{2}}^{\frac{π}{2}} (\sqrt{1 - {cos}^{2} θ} - 1) d θ \\ = & 4 \int_{- \frac{π}{2}}^{\frac{π}{2}} (1 - sin θ) d θ \\ = & 4 [\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d θ - (\int_{- \frac{π}{2}}^{\frac{π}{2}} sin θ d θ)] \\ = & 4 [π - (- \int_{- \frac{π}{2}}^{0} sin θ d θ + \int_{0}^{\frac{π}{2}} sin θ d θ)] \\ = & 4 [π - (- {[- cos θ]}_{- \frac{π}{2}}^{0} + {[- cos θ]}_{0}^{\frac{π}{2}})] \\ = & 4 [π - (- (- cos 0 + cos (- \frac{π}{2})) + (- cos \frac{π}{2} + cos 0))] \\ = & 4 [π - (- (- 1 + 0) + (- 0 + 1))] \\ = & 4 [π - (1 + 1)] \\ = & 4 [π - 2] \\ = & 4 π - 8 \approx 4.566 . \end{array}

Rounded off to two decimal places is 4.57.

So, the correct answer is $\to$ $\underset{_____________}{4.57}$

#Integral Calculus #IIT JAM PYQ 2024 #Mathematics #IIT JAM PYQs

Question

Let S={(x,y,z)∈ℝ3:x2+y2+z2=4,(x−1)2+y2≤1,z≥0}. Then, the surface area of 𝑆 equals _____________ (rounded off to two decimal places).

Solution :-

Let
$S = {(x, y, z) \in ℝ^{3} : x^{2} + y^{2} + z^{2} = 4, {(x - 1)}^{2} + y^{2} \leq 1, z \geq 0} .$
Then, the surface area of 𝑆 equals _____________ (rounded off to two decimal
places).