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Electricity and Magnetism
Mathematical Tools and Fields
01Series Approximations: Keeping One or Two Terms02Integrals as Sums: Along a Line and a Curved Path03Line Integrals: Work as the Integral of F · ds04Closed Line Integrals: Work Around a Loop05Double Integrals: Area, Charge and Volume06Closed Surface Integrals: The Sphere and Its Charge07Electric Flux Through Surfaces and Volume IntegralsProblem set0/10Problem set 20/10MIT problem set0/4
01Field Theory Versus Action at a Distance02Scalar Fields: Contour, Color and Relief Maps03Vector Fields, Sources, Sinks and Divergence04Curl: Measuring the Circulation of a Field05Gravitational, Electric and Magnetic Fields06Drawing Vector Fields: Arrows, Lines and Grass SeedsProblem set0/10Problem set 20/10MIT problem set0/1

Closed Surface Integrals: The Sphere and Its Charge

How does the spherical area element sum to 4πR², and how do you integrate over a closed surface built from pieces?


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Your summary note

    1. 1

      Closed surfaces and the circled double-integral sign

      Define a closed surface as one that completely encloses a volume, give two examples, and record that the loop on ∮ ⁣ ⁣∮\oint\!\!\oint∮∮ marks closure as it does for closed paths.

    2. 2

      Sphere area A=4πR2A = 4\pi R^2A=4πR2 and its area element

      Write the sphere's area as a closed surface integral, write dA=R2sin⁡θ dθ dϕdA = R^2 \sin\theta \, d\theta \, d\phidA=R2sinθdθdϕ with θ\thetaθ polar and ϕ\phiϕ azimuthal, and note that math texts sometimes swap the two angles.

    3. 3

      Uniform charge on a sphere, Q=σ 4πR2Q = \sigma \, 4\pi R^2Q=σ4πR2

      Write QQQ as the closed surface integral of σ dA\sigma \, dAσdA, pull the constant σ\sigmaσ out, replace the remaining integral by the known area, and check the units come out in coulombs.

    4. 4

      Total charge on the two-hemisphere-and-annulus surface

      Sketch the surface, split the closed integral into three open-surface integrals with one density each, then use hemisphere area 2πR22\pi R^22πR2 and annulus area π(R22−R12)\pi(R_2^2 - R_1^2)π(R22​−R12​) to reach QQQ.

    Attempt 1 of 2