Question

the area of the surface of the part of y^2+z^2=x^2 cut off by the cylinder x^2-y^2=a^2 and the planes y=b, y=-b is... ​

Answer 1

Answer:

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

The region of the cylinder is given by the limits 0≤θ≤π, 0≤r≤asinθ 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

where F is the equation of the sphere.

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

A=2a∬1a2−x2−y2−−−−−−−−−−√dxdy

Converting to polar coordinates, we have:

A=2a∫π0∫asin(θ)0ra2−r2−−−−−−√drdθ

Calculating this I get 2πa2. The answer is (2π−4)a2. Where am I going wrong?

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Jul 15 '14 at 13:56

Haresh

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Jul 15 '14 at 14:12

Siminore

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2 Answers

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Given the equations

x2+y2+z2=a2,

and

x2+y2=ay,

we obtain

ay+z2=a2.

Using

xyz===asin(θ)cos(ϕ),asin(θ)sin(ϕ),acos(θ),

we obtain

a2sin(θ)sin(ϕ)+a2cos2(θ)=a2⇒sin(θ)=sin(ϕ)⇒θ=ϕ∨θ=π−ϕ.

For the surface we have

∫dϕ∫dθsin(θ)==∫π/20dϕ∫ϕ0dθsin(θ)+∫πϕ/2dϕ∫π−ϕ0dθsin(θ)2∫π/20dϕ∫ϕ0dθsin(θ).

We can calculate the surface as

4a2∫π/20dϕ∫ϕ0dθsin(θ)===4a2∫π/20dϕ(1−cos(ϕ))4a2(π/2−1)a2(2π−4).