a) Show that the formula for the surface area of a sphere with radius $r$ $r$ is $4 π r^{2}$ $4pir^2$ . b) If a portion of the sphere is removed to form a spherical cap of height $h$ $h$ then then show the curved surface area is $2 π h r^{2}$ $2pihr^2$ ?

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a) Show that the formula for the surface area of a sphere with radius $r$ $r$ is $4 π r^{2}$ $4pir^2$ . b) If a portion of the sphere is removed to form a spherical cap of height $h$ $h$ then then show the curved surface area is $2 π h r^{2}$ $2pihr^2$ ?

a) Show that the formula for the surface area of a sphere with radius $r$ $r$ is $4 π r^{2}$ $4pir^2$ .

b) If a portion of the sphere is removed to form a spherical cap of height $h$ $h$ then then show the curved surface area is $2 π h r^{2}$ $2pihr^2$

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Answer 1 · 2017-04-04T05:47:22

A = int dA

Explanation:

An area element on a sphere has constant radius r, and two angles. One is longitude $ϕ$ $phi$ , which varies from $0$ $0$ to $2 π$ $2pi$ . The other one is the angle with the vertical. To avoid counting twice, that angle only varies between $0$ $0$ and $π$ $pi$ .

So the area element is $d A = r d θ r sin θ d ϕ = r^{2} sin θ d θ d ϕ$ $dA = r d theta r sin theta d phi = r^2 sin theta d theta d phi$
Integrated over the whole sphere gives
$A = \int_{0}^{π} sin θ d θ \int_{0}^{2 π} d ϕ r^{2} = - cos (θ) ∣_{0}^{π} 2 π r^{2} = 4 π r^{2}$ $A = int_0^pi sin theta d theta int_0^(2pi) dphi r^2 = -cos(theta)|_0^(pi) 2 pi r^2 = 4 pi r^2$

In part b, $cos (θ)$ $cos(theta)$ varies between $\frac{a}{r}$ $a/r$ and $\frac{b}{r}$ $b/r$ which is such that $b - a = h$ $b-a=h$
Then $A = cos θ ∣_{b}^{a} 2 π r^{2} = (\frac{b}{r} - \frac{a}{r}) 2 π r^{2} = (\frac{h}{r}) 2 π r^{2} = 2 π r h$ $A= cos theta|_b^a 2pi r^2 = (b/r - a/r) 2 pi r^2 = (h/r) 2pi r^2 = 2 pi r h$

Note: Every derivation I found of this result uses cylindrical coordinates and is far more involved than this one. Can someone else check?

Answer 2 · 2017-04-04T10:43:25

a) $A = 4 π r^{2}$ $A = 4pir^2$
b) $A = 2 π h r^{2}$ $A = 2pihr^2$

Explanation:

It is easier to use Spherical Coordinates, rather than Cylindrical or rectangular coordinates. This solution looks long because I have broken down every step, but it can be computer in just a few lines of calculation

With spherical coordinates, we can define a sphere of radius $r$ $r$ by all coordinate points where $0 \leq ϕ \leq π$ $0 le phi le pi$ (Where $ϕ$ $phi$ is the angle measured down from the positive $z$ $z$ -axis), and $0 \leq θ \leq 2 π$ $0 le theta le 2pi$ (just the same as it would be polar coordinates), and $ρ = r$ $rho=r$ ).

The Jacobian for Spherical Coordinates is given by $J = ρ^{2} sin ϕ$ $J=rho^2 sin phi$

And so we can calculate the surface area of a sphere of radius $r$ $r$ using a double integral:

$A = int int_R \ \ dS \ \ \$

where $R={(x,y,z) in RR^3 | x^2+y^2+z^2 = r^2 }$

$:. A = int_0^pi \ int_0^(2pi) \ r^2 sin phi \ d theta \ d phi$

If we look at the inner integral we have:

$int_0^(2pi) \ r^2 sin phi \ d theta = r^2sin phi \ int_0^(2pi) \ d theta$
$" " = r^2sin phi [ \ theta \ ]_0^(2pi)$
$" " = (r^2sin phi) (2pi-0)$
$" " = 2pir^2 sin phi$

So our integral becomes:

$A = int_0^pi \ 2pir^2 sin phi \ d phi$
$\ \ \ = -2pir^2 { cos phi ]_0^pi$
$\ \ \ = -2pir^2 (cospi-cos0)$
$\ \ \ = -2pir^2 (-1-1)$
$\ \ \ = -2pir^2 (-2)$
$\ \ \ = 4pir^2 \ \ \$ QED

For the area of the portion of a sphere we have a similar set-up:

By trigonometry $cos phi = h/r => phi = arccos(h/r)$ , and so we must restrict $phi$ to $arccos(h/r) le phi le pi/2$ , which gives us:

$A = int_arccos(h/r)^(pi/2) \ int_0^(2pi) \ r^2 sin phi \ d theta \ d phi$
$\ \ \ = int_arccos(h/r)^(pi/2) \ (r^2sin phi)(2pi-0) \ d phi$
$\ \ \ = int_arccos(h/r)^(pi/2) \ 2pir^2sin phi \ d phi$
$\ \ \ = -2pir^2[cosphi]_arccos(h/r)^(pi/2)$
$\ \ \ = -2pir^2(cos(pi/2)-cos(arccos(h/r)))$
$\ \ \ = -2pir^2(0-h/r)$
$\ \ \ = -2pir^2(-h/r)$
$\ \ \ = 2pihr^2 \ \ \$ QED

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a) Show that the formula for the surface area of a sphere with radius $r$ $r$ is $4 π r^{2}$ $4pir^2$ . b) If a portion of the sphere is removed to form a spherical cap of height $h$ $h$ then then show the curved surface area is $2 π h r^{2}$ $2pihr^2$ ?