Question

Calculate the surface integral \iint_M (\nabla \times \mathbf{F}) \cdot d\mathbf{S} where $M$ is the hemisphere $x^2 + y^2 + z^2 = 4$, $x \ge 0$, with the normal in the direction of the positive x direction, and $\mathbf{F} = (x^6, 0, y^2)$.\newline Begin by writing down the \"standard\" parametrization of $\partial M$ as a function of the angle $\theta$ (denoted by \"t\" in your answer)\newline $x = 2\cos t \sin t$, $y = 2\sin t \sin t$, $z = 2\cos t$.\newline $\int_{\partial M} \mathbf{F} \cdot d\mathbf{s} = \int_0^{2\pi} f(\theta) \, d\theta$, where\newline $f(\theta) = \cos^7 t$ (use \"t\" for theta).\newline The value of the integral is $\frac{6552\pi}{315}$

          Calculate the surface integral \iint_M (\nabla \times \mathbf{F}) \cdot d\mathbf{S} where $M$ is the hemisphere $x^2 + y^2 + z^2 = 4$, $x \ge 0$, with the normal in the direction of the positive x direction, and $\mathbf{F} = (x^6, 0, y^2)$.\newline Begin by writing down the \"standard\" parametrization of $\partial M$ as a function of the angle $\theta$ (denoted by \"t\" in your answer)\newline $x = 2\cos t \sin t$, $y = 2\sin t \sin t$, $z = 2\cos t$.\newline $\int_{\partial M} \mathbf{F} \cdot d\mathbf{s} = \int_0^{2\pi} f(\theta) \, d\theta$, where\newline $f(\theta) = \cos^7 t$ (use \"t\" for theta).\newline The value of the integral is $\frac{6552\pi}{315}$

Calculate the surface integral (∇×𝐅) ·d𝐒 where M is the hemisphere x^2 + y^2 + z^2 = 4, x ≥ 0, with the normal in the direction of the positive x direction, and 𝐅 = (x^6, 0, y^2).Begin by writing down the s̈tandardp̈arametrization of ∂ M as a function of the angle θ (denoted by ẗïn your answer)x = 2cos t sin t, y = 2sin t sin t, z = 2cos t.∫∂ M𝐅· d𝐬 = ∫0^2π f(θ) dθ, wheref(θ) = cos^7 t (use ẗf̈or theta).The value of the integral is (6552π)/(315)

Added by Patricia G.

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