Use stokes' theorem to evaluate s curl f · ds. F(x, y, z) = x2 sin(z)i + y2j + xyk, s is the part of the paraboloid z = 4 − x2 − y2 that lies above the xy-plane, oriented upward.

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Answer 1

Stokes' theorem says that the surface integral of the curl of the vector field F across the surface S is equal to the line integral of F along the boundary of S.

In this case, the boundary is a circle in the x,y-plane with radius 2. (To see this, set z = 0 and rearrange the equation for the paraboloid as x² + y² = 4.)

Then

[tex]\displaystyle \iint_S (\nabla\times\vec F)\cdot d\vec s = \int_{\partial S} \vec F \cdot d\vec r[/tex]

where

[tex]\vec r = 2\cos(t) \,\vec\imath + 2\sin(t) \,\vec\jmath[/tex]

parameterizes the circular boundary of S with 0 ≤ t ≤ 2π. Then

[tex]d\vec r = \dfrac{d\vec r}{dt}\,\dt = (-2\sin(t)\,\vec\imath + 2\cos(t)\,\vec\jmath)\,dt[/tex]

and the integral becomes

[tex]\displaystyle \iint_S (\nabla\times\vec F)\cdot d\vec s = \int_{\partial S} \vec F \cdot d\vec r[/tex]

[tex]\displaystyle \iint_S (\nabla\times\vec F)\cdot d\vec s = \int_0^{2\pi} \vec F(\vec r(t)) \cdot \frac{d\vec r}{dt}\,dt[/tex]

[tex]\displaystyle \iint_S (\nabla\times\vec F)\cdot d\vec s \\= \int_0^{2\pi} ((2\cos(t))^2 \sin(0)\,\vec\imath + (2\sin(t))^2\,\vec\jmath + (2\cos(t))(2\sin(t))\,\vec k) \cdot(-2\sin(t)\,\vec\imath + 2\cos(t)\,\vec\jmath)\,dt[/tex]

[tex]\displaystyle \iint_S (\nabla\times\vec F)\cdot d\vec s = \int_0^{2\pi} 8\sin^2(t)\cos(t) \, dt[/tex]

[tex]\displaystyle \iint_S (\nabla\times\vec F)\cdot d\vec s = \int_0^{2\pi} 8\sin^2(t) \, d(\sin(t))[/tex]

[tex]\displaystyle \iint_S (\nabla\times\vec F)\cdot d\vec s = \frac83 (\sin^3(2\pi)-\sin^3(0)) = \boxed{0}[/tex]


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Answer:

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Explanation:

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Answer:

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Answer:

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The answer to your question is: D.) 9x²-11x-12y²

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Which set of equations is enough information to prove that lines c and d are parallel lines cut by transversal p? m∠1 = 81° and m∠2 = 99° m∠3 = 99° and m∠4 = 99° m∠2 = 99° and m∠4 = 99° m∠4 = 81° and m∠1 = 81°.

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The set of equation that is enough information to prove that lines c and d are parallel lines cut by transversal p is m∠2 = 99° and m∠4 = 99°. The c.

What are transverse lines?

A transversal line is a line that crosses two or more other lines at different locations.  Line l meets a and b in the image below at two different locations, P and Q. Line l is the transversal line as a result.

When a transversal cuts two lines so that their subsequent interior angles are supplementary, the lines are said to be parallel. Two lines are parallel to one another if they are both parallel to the same line. When the marked consecutive internal angles are supplementary, lines m and n are parallel.

Therefore, the correct option is c, m∠2 = 99° and m∠4 = 99°.

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The question is incomplete. Your most probably complete question is given below: the image is added.

The set of equation that is enough information to prove that lines c and d are parallel lines cut by transversal p is m∠2 = 99° and m∠4 = 99°. The c.

What are transverse lines?

A transversal line is a line that crosses two or more other lines at different locations.  Line l meets a and b in the image below at two different locations, P and Q. Line l is the transversal line as a result.

When a transversal cuts two lines so that their subsequent interior angles are supplementary, the lines are said to be parallel. Two lines are parallel to one another if they are both parallel to the same line. When the marked consecutive internal angles are supplementary, lines m and n are parallel.

Therefore, the correct option is c, m∠2 = 99° and m∠4 = 99°.The set of equation that is enough information to prove that lines c and d are parallel lines cut by transversal p is m∠2 = 99° and m∠4 = 99°. The c.

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Answer:

It lets people practice their faith in their own way and

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Explanation:

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