Question

Verify stokes' theorem for the helicoid ψ(r,θ)=⟨rcosθ,rsinθ,θ⟩ where (r,θ) lies in the rectangle [0,1]×[0,π/2], and f is the vector field f=⟨6z,8x,8y⟩. first, compute the surface integral: ∬m(∇×f)⋅ds=∫ba∫dcf(r,θ)drdθ, where a= , b= , c= , d= , and f(r,θ)= (use "t" for theta). finally, the value of the surface integral is . next compute the line integral on that part of the boundary from (1,0,0) to (0,1,π/2). ∫cf⋅dr=∫bag(θ)dθ, where a= , b= , and g(θ)=

Answer 1

$\mathbf f(x,y,z)=\langle6z,8x,8y\rangle\implies\nabla\times\mathbf f(x,y,z)=\langle8,6,8\rangle$

$\psi(r,\theta)=\langle r\cos\theta,r\sin\theta,\theta\rangle$
$\mathrm d\mathbf S=\dfrac{\psi_r\times\psi_\theta}{\left\|\psi_r\times\psi_\theta\right\|}\left\|\psi_r\times\psi_\theta\right\|\,\mathrm dr\,\mathrm d\theta=\langle\sin \theta,-\cos \theta,r\rangle\,\mathrm dr\,\mathrm d\theta$

$\displaystyle\iint_S\nabla\times\mathbf f\cdot\mathrm d\mathbf S=\int_{\theta=0}^{\theta=\pi/2}\int_{r=0}^{r=1}\langle8,6,8\rangle\cdot\langle\sin\theta,-\cos\theta,r\rangle\,\mathrm dr\,\mathrm d\theta$
$=\displaystyle\int_{\theta=0}^{\theta=\pi/2}\int_{r=0}^{r=1}(8r-6\cos\theta+8\sin\theta)\,\mathrm dr\,\mathrm d\theta=2+2\pi$

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$\mathbr r(\theta)=\langle\cos\theta,\sin\theta,\theta\rangle$

$\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r=\int_{\theta=0}^{\theta=\pi/2}\mathbf f(\cos\theta,\sin\theta,\theta)\cdot\langle-\sin\theta,\cos\theta,1\rangle\,\mathrm d\theta$
$=\displaystyle\int_0^{\pi/2}\langle6\theta,8\cos\theta,8\sin\theta\rangle\cdot\langle-\sin\theta,\cos\theta,1\rangle\,\mathrm d\theta$
$=\displaystyle\int_0^{\pi/2}(8\cos^2\theta+(8-6t)\sin\theta)\,\mathrm d\theta=2+2\pi$