Proof that surface area of sphere is
Answers
To prove that the surface area of a sphere of radius rr is 4 \pi r^24πr
2
, one straightforward method we can use is calculus. We first have to realize that for a curve parameterized by x(t)x(t) and y(ty(t), the arc length is
S = \int_a^b \sqrt{ \left(\frac{dy}{dt}\right)^2 + \left( \frac{dx}{dt}\right)^2 } \, dt.
S=∫
a
b
(
dt
dy
)
2
+(
dt
dx
)
2
dt.
From this we can derive the formula for the surface area of the solid obtained by rotating this about the xx-axis. This turns out to be
A = 2\pi \int_a^b y\sqrt{ \left(\frac{dy}{dt}\right)^2 + \left( \frac{dx}{dt}\right)^2 } \, dt .
A=2π∫
a
b
y
(
dt
dy
)
2
+(
dt
dx
)
2
dt.
We can obtain a sphere by revolving half a circle about the xx-axis. This circle can be parameterized as x(t)=r\cos(t)x(t)=rcos(t) and y(t) = r\sin(t)y(t)=rsin(t) for 0 \leq t \leq \pi0≤t≤π. From this, we get
\frac{dx}{dt} = -r\sin(t), \quad \frac{dy}{dt} = r\cos(t) .
dt
dx
=−rsin(t),
dt
dy
=rcos(t).
Substituting in our equations for surface area gives
\begin{aligned} A &= 2\pi \int_0^\pi r\sin(t)\sqrt{ \big(-r\sin(t)\big)^2 + \big( r\cos(t) \big)^2 } \ dt \\ &= 2\pi \int_0^\pi r\sin(t)\sqrt{ r^2\big(\sin(t)^2 + \cos(t)^2 \big) } \ dt \\ &= 2\pi \int_0^\pi r^2 \sin(t) \ dt \\ &= 2\pi r^2 \int_0^\pi\sin(t) \ dt \\ &= 4 \pi r^2. \ _\square \end{aligned}
A
=2π∫
0
π
rsin(t)
(−rsin(t))
2
+(rcos(t))
2
dt
=2π∫
0
π
rsin(t)
r
2
(sin(t)
2
+cos(t)
2
)
dt
=2π∫
0
π
r
2
sin(t) dt
=2πr
2
∫
0
π
sin(t) dt
=4πr
2
.
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