Evaluate the line integral ∫cf⋅dr, where f(x,y,z)=−4xi−3yj+4zk and c is given by the vector function r(t)=⟨sint,cost,t⟩, 0≤t≤3π/2.

[tex]\displaystyle\int_C\mathbf f\cdot\mathrm d\mathbf r[/tex]
[tex]=\displaystyle\int_{t=0}^{t=3\pi/2}\mathbf f(\sin t,\cos t,t)\cdot\langle\cos t,-\sin t,1\rangle\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^{3\pi/2}\langle-4\sin t,-3\cos t,4t\rangle\cdot\langle\cos t,-\sin t,1\rangle\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^{3\pi/2}(4t-\cos t\sin t)\,\mathrm dt=\dfrac{9\pi^2-1}2[/tex]

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yoodyannapolis yoodyannapolis · Answer 2 · 2021-11-29T07:00:54+01:00

The line integral [tex]\int\limits_c {f} .\, dr[/tex] where [tex]f(x,y,z)=-4xi-3yj+4zk[/tex] and c is given by the vector function [tex]r(t)=(sint,cost,t), 0\leq t\leq \frac{3\pi}{2}[/tex] is [tex]\frac{9\pi^{2}-1 }{2}[/tex].

We have [tex]f(x,y,z)=-4xi-3yj+4zk[/tex] and [tex]r(t)=(sint,cost,t), 0\leq t\leq \frac{3\pi}{2}[/tex]

Now, [tex]\int\limits^\frac{3\pi}{2} _0 {f} .\, dr \\=\int\limits^\frac{3\pi}{2} _0 {f(sint,cost,t)}(cost,-sint,1) .\, dt \\=\int\limits^\frac{3\pi}{2} _0 {(-4sint,-3cost,+4t)}(cost,-sint,1) .\, dt\\=\int\limits^\frac{3\pi}{2} _0 {(4t,-costsint) .\, dt\\=\int\limits^\frac{3\pi}{2} _0 {(4t,-\frac{1}{2} .2costsint) .\, dt\\=\int\limits^\frac{3\pi}{2} _0 {(4t,-\frac{1}{2} .sin2t) .\, dt \\=\frac{9\pi^{2}-1 }{2}[/tex]

Therefore, the line integral [tex]\int\limits_c {f} .\, dr[/tex] where [tex]f(x,y,z)=-4xi-3yj+4zk[/tex] and c is given by the vector function [tex]r(t)=(sint,cost,t), 0\leq t\leq \frac{3\pi}{2}[/tex] is [tex]\frac{9\pi^{2}-1 }{2}[/tex].

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