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We compute the integral of f(x,y,z)=xz over the upper hemisphere of radius 3 twice: once in spherical coordinates and once in cylindrical coordinates. The result for both will be 0, because the integrand is odd symmetric in x. However, I think for those of you who need practice with trigonometry, it's still a worthy exercise to do in detail. We begin with spherical coordinates, setting up our parametrization for S with ρ constant at 3. Our coordinates are x=3cos(u)sin(v), y=3sin(u)sin(v), and z=3cos(v). The bounds for u span from 0 to 2pi and for v from 0 to π/2. We express f in these coordinates as 9cos(u)cos(v)sin(v). Next, we calculate the necessary cross product of the derivatives of our parametrization with respect to u and v, simplifying the resulting vectors and their magnitudes. After incorporating a sine factor from the sphere's geometry, the length of the cross product vector simplifies to 9sin(v), which greatly simplifies the integral setup. The double integral, extended over the bounds of u and v, involves the product of our simplified integrand and the vector magnitude, resulting in zero due to the integration of cos(u) over its complete period, which sums to zero. We then repeat the process using cylindrical coordinates, focusing on how the upper hemisphere projects a circular shadow on the xy-plane. We set x and y based on the radial and angular coordinates, and z from the sphere's equation adapted to these coordinates. The cross product of the derivatives in cylindrical coordinates involves similar steps, leading to an integral setup that also reduces to zero because the integration of cos(v) over 0 to 2π also sums to zero. #mathematics #math #calculus #multivariablecalculus #surfaceintegral #iitjammathematics #calculus3 #sphericalcoordinates #cylindricalcoordinates #mathtutorial
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