Use Stokes' theorem to evaluateV XF- î ds where F = (9y?z, 4xz, 2x²y²) and S is the paraboloid z = x² + y? inside the cylinder x? + y? = 1, oriented upward.

Step:-1 Stoke's Theorem:- ∬s∇×F→.n dS=∫F.dr Now, given that F→=9y2z, 4xz, 2x2y2 and paraboloid…

Answered: Use Stokes' theorem to evaluate V ×F· î…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Use Stokes' theorem to evaluate
V XF- î ds where F = (9y?z, 4xz, 2x²y²) and S is the paraboloid z = x² + y? inside the cylinder x? + y? = 1, oriented upward.

Expert Solution

Step 1

Step:-1

Stoke's Theorem:-

$\iint_{s} \nabla \times \vec{F} . n d S = \int F . d r$

Now, given that $\vec{F} = <9 y^{2} z, 4 x z, 2 x^{2} y^{2}>$ and paraboloid $z = x^{2} + y^{2}$ inside the cylinder $x^{2} + y^{2} = 1 \Rightarrow z = 1$

So, boundary curve is circle $x^{2} + y^{2} = 1$ with radius 1, in the plane z=1.

Take a vector $\vec{r} (x, y, z) = x i + y j + z k$ , parameterization this vector so,

Take $x = \cos t, y = \sin t a n d z = 1$ then

$\vec{r} (t) = \cos t i + \sin t j + k \frac{d r}{d t} = - \sin t i + \cos t j + 0 \frac{d r}{d t} = <- \sin t, \cos t, 0>$

and $0 \leq t \leq 2 π$

$\vec{F} = <9 y^{2} z, 4 x z, 2 x^{2} y^{2}> \vec{F} (t) = <9 \sin^{2} t, 4 \cos t, 2 \cos^{2} t \sin^{2} t> (a s x = \cos t, y = \sin t, z = 1)$

Step:-2

$\int F . d r = \int_{0}^{2 π} F (t) . \frac{d r}{d t} . d t = \int_{0}^{2 π} <9 \sin^{2} t, 4 \cos t, 2 \cos^{2} t \sin^{2} t> <- \sin t, \cos t, 0> d t \int F . d r = = \int_{0}^{2 π} (- 9 \sin^{3} t + 4 \cos^{2} t) d t N o t e : - \cos (2 t) = 2 \cos^{2} t - 1 \Rightarrow 2 \cos^{2} t = \cos (2 t) + 1 a n d \sin (3 t) = 3 \sin t - 4 \sin^{3} t \Rightarrow \sin^{3} t = \frac{3 \sin t - \sin (3 t)}{4} \int F . d r = \int_{0}^{2 π} (- 9 \frac{3 \sin t - \sin (3 t)}{4} + 2 (\cos (2 t) + 1)) d t = \int_{0}^{2 π} (\frac{- 27}{4} \sin t + \frac{9}{4} \sin (3 t) + 2 \cos (2 t) + 2) d t \int F . d r = \frac{27}{4} {[\cos t]}_{0}^{2 π} + \frac{(- 9)}{12} {[\cos t]}_{0}^{2 π} + {[\sin t]}_{0}^{2 π} + 2 {[t]}_{0}^{2 π} \int F . d r = 0 + 0 + 0 + 4 π = 4 π \int F . dr = 4 π$

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