Mathematics: Let f = (z − y) i + (x − z) j + (y − x) k . let c be the rectangle of width 2 and length 5 centered at (7, 7, 7) on the plane x + y + z = 21, oriented clockwise when viewed from the origin. (a) find curlf . curlf = ⟨2,2,2⟩ (b) use stokes theorem to find f · dr c . f · dr c = −60 √3

Let f = (z − y) i + (x − z) j + (y − x) k . let

Ответ:

hfkdnd

10.03.2022

I suppose the rectangle's width and length refer to the dimensions of the rectangle that the plane x+y+z=21 projects onto the x,y plane, so that $2|x-7|\le2$ and $2|y-7|\le5$ , or equivalently $6\le x\le8$ and $\dfrac92\le y\le\dfrac{19}2$ .

The part of the plane bounded by can be parameterized by

$\vec s(x,y)=x\,\vec\imath+y\,\vec\jmath+(21-x-y)\,\vec k$

with the bounds on and listed above. Call this surface . Take the normal vector to to be

$\vec s_x\times\vec s_y=\vec\imath+\vec\jmath+\vec k$

a. Your answer for the curl is correct, $\nabla\times\vec F(x,y,z)=2(\vec\imath+\vec\jmath+\vec k)$ .

b. By Stokes' theorem,

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S$

$=\displaystyle\int_{9/2}^{19/2}\int_6^82(\vec\imath+\vec\jmath+\vec k)\cdot(\vec\imath+\vec\jmath+\vec k)\,\mathrm dx\,\mathrm dy$

$=\displaystyle6\int_{9/2}^{19/2}\int_6^8\mathrm dx\,\mathrm dy=\boxed{60}$

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Ответ:

animaljamissofab

03.01.2022

Mathematics

Hello,

Answer A

Yes, as long as each number in the Pythagorean triple is multiplied by the same whole number.

Step-by-step explanation:

(a,b,c) is a Pythagorien triple : c²=a²+b²

and k is a naturel.

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Let f = (z − y) i + (x − z) j + (y − x) k . let c be the rectangle of width 2 and length 5 centered at (7, 7, 7) on the plane x + y + z = 21, oriented clockwise when viewed from the origin. (a) find curlf . curlf = ⟨2,2,2⟩ (b) use stokes' theorem to find f · dr c . f · dr c = −60 √3

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