(Vector Calculus) Help regarding area element notation

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SUMMARY

The area element of a sphere in spherical coordinates is expressed as dA = r^2 sin(φ) dθ dφ. A substitution method allows for the transformation of the integral ∫₀^π sin(φ) dφ into ∫₋₁¹ d(cos(φ)), where u = cos(φ) and du = -sin(φ) dφ. This substitution is valid as it maintains the limits of integration, resulting in the equality ∫₀^π sin(φ) dφ = ∫₋₁¹ d(cos(φ)). This method is essential for simplifying surface area calculations in vector calculus.

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Homework Statement


The area element of a sphere in spherical coordinates is given as following
[itex]dA = r^2 \sin(\phi)\; d \theta \; d \phi[/itex]​

using the notation in the following figure:
SphericalCoordinates_1201.gif


However, while going through some E&M books I ran into the following notation

[itex]Surface \; Area = r^2 \; \int_{-1} ^1 d \cos(\phi) \; \int_0^{2 \pi}d \theta \; = 4 \pi r^2[/itex]​

My question is how can we replace [itex]\int_{0} ^\pi \sin(\phi) \; d \phi[/itex] with [itex]\int_{-1} ^1 d \cos(\phi)[/itex]
 
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That's basically a substitution. If you let [itex]u= cos(\phi)[/itex] then [itex]du= d(cos(\phi)= -sin(\phi)d\phi[/itex]. Also, when [itex]\phi= 0[/itex], [itex]cos(\phi= 1[/itex] and when [itex]\phi= \pi[/itex], [itex]cos(\phi)= -1[/itex].

With that substitution, [itex]\int_0^\pi sin(\phi)d\phi= \int_1^{-1} -du[/itex] and, of course, swapping the limits of integration multiplies the integral by -1:
[itex]\int_0^\pi sin(\phi)d\phi= \int_1^{-1} -du= \int_{-1}^1 du= \int_{-1}^1 d(cos(\phi))[/itex]
 

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