Integration over surfaces

Gauss' law is a very powerful method to determine the electric field due to a distribution of charges. The mathematical expression for Gauss' law is:

S E d A = Q e n c ϵ 0 \int_{S} \vec{E} \cdot \vec{dA}=\frac{Q_{enc}}{\epsilon_0}

where S S is a surface, E \vec{E} is the electric field vector, d A \vec{dA} is the infinitesimal area element, Q e n c Q_{enc} is the charge enclosed by S S and ϵ 0 \epsilon_0 is a constant.

We now turn to the notion of integration over a specified surface. One can think about definite integration over a line interval, i.e. a b f ( x ) d x \int_a^b f(x) dx , as the area under the curve defined by f ( x ) f(x) . Similarly, the integral over a surface of a function can be thought of as the volume under the 2-d graph of the function. Using this definition, what is the integral

S x d A \int_S x dA

where the surface S S is the square in the x y xy -plane with corners at ( 0 , 0 ) , ( 1 , 0 ) , ( 0 , 1 ) , ( 1 , 1 ) (0,0), (1,0),(0,1),(1,1) .


This problem is part of David's set on Gauss' Law .


The answer is 0.5.

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1 solution

Discussions for this problem are now closed

We have like d A = d x d y dA=dxdy , where x x and y y runs from 0 0 to 1 1 .

So, s x d A = x d x d y = 0 1 x d x 0 1 d y = 1 2 \int_sxdA = \int\int xdxdy = \int_0^1xdx\int_0^1dy=\boxed{\frac{1}{2}}

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