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My name is Ann [436]
3 years ago
15

Evaluate the surface integral S F · dS for the given vector field F and the oriented surface S. In other words, find the flux of

F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = x i − z j + y k S is the part of the sphere x2 + y2 + z2 = 49 in the first octant, with orientation toward the origin
Mathematics
1 answer:
Tomtit [17]3 years ago
4 0

Apparently my answer was unclear the first time?

The flux of <em>F</em> across <em>S</em> is given by the surface integral,

\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S

Parameterize <em>S</em> by the vector-valued function <em>r</em>(<em>u</em>, <em>v</em>) defined by

\mathbf r(u,v)=7\cos u\sin v\,\mathbf i+7\sin u\sin v\,\mathbf j+7\cos v\,\mathbf k

with 0 ≤ <em>u</em> ≤ π/2 and 0 ≤ <em>v</em> ≤ π/2. Then the surface element is

d<em>S</em> = <em>n</em> • d<em>S</em>

where <em>n</em> is the normal vector to the surface. Take it to be

\mathbf n=\dfrac{\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}}{\left\|\frac{\partial\mathbf r}{\partial v}\times\frac{\partial\mathbf r}{\partial u}\right\|}

The surface element reduces to

\mathrm d\mathbf S=\mathbf n\,\mathrm dS=\mathbf n\left\|\dfrac{\partial\mathbf r}{\partial u}\times\dfrac{\partial\mathbf r}{\partial v}\right\|\,\mathrm du\,\mathrm dv

\implies\mathbf n\,\mathrm dS=-49(\cos u\sin^2v\,\mathbf i+\sin u\sin^2v\,\mathbf j+\cos v\sin v\,\mathbf k)\,\mathrm du\,\mathrm dv

so that it points toward the origin at any point on <em>S</em>.

Then the integral with respect to <em>u</em> and <em>v</em> is

\displaystyle\iint_S\mathbf F\cdot\mathrm d\mathbf S=\int_0^{\pi/2}\int_0^{\pi/2}\mathbf F(x(u,v),y(u,v),z(u,v))\cdot\mathbf n\,\mathrm dS

=\displaystyle-49\int_0^{\pi/2}\int_0^{\pi/2}(7\cos u\sin v\,\mathbf i-7\cos v\,\mathbf j+7\sin u\sin v\,\mathbf )\cdot\mathbf n\,\mathrm dS

=-343\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}\cos^2u\sin^3v\,\mathrm du\,\mathrm dv=\boxed{-\frac{343\pi}6}

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The value of c for which the considered trinomial becomes perfect square trinomial is: 20 or -20

<h3>What are perfect squares trinomials?</h3>

They are those expressions which are found by squaring binomial expressions.

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Answer:

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  2. (x, y) = (1, 2)

Step-by-step explanation:

A nice graphing calculator app makes these trivially simple. (See the first two attachments.) It is available for phones, tablets, and as a web page.

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The usual methods of solving a system of equations involve <em>elimination</em> or <em>substitution</em>.

There is another method that is relatively easy to use. It is a variation of "Cramer's Rule" and is fully equivalent to <em>elimination</em>. It makes use of a formula applied to the equation coefficients. The pattern of coefficients in the formula, and the formula itself are shown in the third attachment. I like this when the coefficient numbers are "too messy" for elimination or substitution to be used easily. It makes use of the equations in standard form.

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1. In standard form, your equations are ...

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Then the solution is ...

  x=\dfrac{-3(25)-(2)(-9)}{-3(5)-(2)(2)}=\dfrac{-57}{-19}=3\\\\y=\dfrac{-9(5)-(25)(2)}{-19}=\dfrac{-95}{-19}=5\\\\(x,y)=(3,5)

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<em>Note on Cramer's Rule</em>

The equation you will see for Cramer's Rule applied to a system of 2 equations in 2 unknowns will have the terms in numerator and denominator swapped: ec-bf, for example, instead of bf-ec. This effectively multiplies both numerator and denominator by -1, so has no effect on the result.

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