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morpeh [17]
3 years ago
11

Calculate the flux of the vector field F⃗ (x,y,z)=(exy+9z+4)i⃗ +(exy+4z+9)j⃗ +(9z+exy)k⃗ through the square of side length 3 wit

h one vertex at the origin, one edge along the positive y-axis, one edge in the xz-plane with x≥0 and z≥0, oriented downward with normal n⃗ =i⃗ −k
Mathematics
1 answer:
ikadub [295]3 years ago
3 0

The square (call it S) has one vertex at the origin (0, 0, 0) and one edge on the y-axis, which tells us another vertex is (0, 3, 0). The normal vector to the plane is \vec n=\vec\imath-\vec k, which is enough information to figure out the equation of the plane containing S:

(x\,\vec\imath+y\,\vec\jmath+z\,\vec k)\cdot(\vec\imath-\vec k)=0\implies x-z=0\implies z=x

We can parameterize this surface by

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

for 0\le x\le\frac3{\sqrt2} and 0\le y\le3. Then the flux of \vec F, assumed to be

\vec F(x,y,z)=(e^{xy}+9z+4)\,\vec\imath+(e^{xy}+4z+9)\,\vec\jmath+(9ze^{xy})\,\vec k,

is

\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\iint_S\vec F(\vec s(x,y))\cdot\vec n\,\mathrm dx\,\mathrm dy

=\displaystyle\int_0^3\int_0^{3/\sqrt2}\left((4+e^{xy}+9x)\,\vec\imath+(9+e^{xy}+4x)\,\vec\jmath+(e^{xy}+9x)\,\vec k\right)\cdot(\vec\imath-\vec k)\,\mathrm dx\,\mathrm dy

=\displaystyle\int_0^3\int_0^{3/\sqrt2}4\,\mathrm dx\,\mathrm dy=\boxed{18\sqrt2}

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a) The 99% confidence interval would be given by (5004.168;12603.832)

b) For this case we can use the lower bound for the confidence interval as estimation, on this case 5004.168. But we need to take in count that this value it's with 99% of confidence and 1% of significance and 1% of probability to commit type Error I

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A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

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\bar X =9004 represent the sample mean for the sample  

\mu population mean (variable of interest)

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n=20 represent the sample size  

2) Part a

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=20-1=19

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,19)".And we see that t_{\alpha/2}=2.86

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9004-2.86\frac{5629}{\sqrt{20}}=5004.168    

9004+2.86\frac{5629}{\sqrt{20}}=12603.832

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