Question

Find the area of the cap cut from the Hemisphere x2 + y2 + z2 = 2, by the cylinder x2 + y2 = 1 ('z' is greater than or equal to 0).

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Answer 1

Answer:

The cylinder is given by the equation x2+(y−a2)2=(a2)2x2+(y−a2)2=(a2)2.

The region of the cylinder is given by the limits 0≤θ≤π0≤θ≤π, 0≤r≤asinθ0≤r≤asin⁡θ in polar coordinates.

We need to only calculate the surface from a hemisphere and multiply it by two. By implicit functions we have:

A=2∬(∂F∂x)2+(∂F∂y)2+(∂F∂z)2−−−−−−−−−−−−−−−−−−−−√∣∣∂F∂z∣∣dA

A=2∬(∂F∂x)2+(∂F∂y)2+(∂F∂z)2|∂F∂z|dA

where FF is the equation of the sphere.

Plugging in the expressions and simplifying (z≥0)z≥0), we get:

A=2a∬1a2−x2−y2−−−−−−−−−−√dxdy

A=2a∬1a2−x2−y2dxdy

Converting to polar coordinates, we have:

A=2a∫π0∫asin(θ)0ra2−r2−−−−−−√drdθ

A=2a∫0π∫0asin⁡(θ)ra2−r2drdθ

Calculating this I get 2πa22πa2. The answer is (2π−4)a2(2π−4)a2.

Answer 2

Answer:

Final Answer.

Step-by-step explanation:

Given,

Hemisphere by the cylinder where z is greater than equal to 0.

We need to calculate the surface area of the cap of the hemisphere.

K = $x^{2}$+$y^{2}$+$z^{2}$ = 2 by the cylinder, k = $x^{2}$+$y^{2}$ = 1

Now, let the surface be S and it projects x-y plane onto the circular disk

$x^{2}$+$y^{2}$$\leq$s(R).

We also see that p=k is normal to R.

Now,

f(x, y, z) = $x^{2}$+$y^{2}$+$z^{2}$

Grad(f) = 2xi+2yj+2zk

|Grad(f)| = $2\sqrt{2}$

Using the formula for the surface area we get,

Surface Area = $\int\limits \int\limits_R {|Grad(f)/|Grad(f).p|} \, dA$

= $\int\limits \, \int\limits_R {\sqrt{2}/2z } \, dx dy$

= $\sqrt{2} \int\limits \int\limits_R {1/z} \, dx dy$

Now,

z = $\sqrt{2 - x^{2} -y^{2} }$

Therefore, S = $\sqrt{2} \int\limits \, \int\limits_R {1/\sqrt{2-x^{2} -y^{2} } } \, dx dy$

Now, converting it to polar co-ordinate:

Substituting 2-$r^{2}$ = $t^{2}$,

⇒ -2r.dr = 2t.dt

S = $\sqrt{2}\int\limits^p_q \int\limits^a_b {-t.dt/t} \, d(theta)$  where, a=1 and b=$\sqrt{2}$ & p=2π and q=0

S = $\sqrt{2}\int\limits^a_b {\sqrt{2}-1 } \, d(theta)$     where, a=2π and b=0

S = $\sqrt{2}$ ($\sqrt{2}$-1)2π

S = 2π (2-$\sqrt{2}$)

Hence, the required surface area is 2π (2-$\sqrt{2}$).

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