By the divergence theorem, the surface integral over
![S_2](https://tex.z-dn.net/?f=S_2)
is
![\displaystyle\iint_{S_2}\mathbf F\cdot\mathrm dS=\iiint_R\nabla\cdot\mathbf F\,\mathrm dR](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_%7BS_2%7D%5Cmathbf%20F%5Ccdot%5Cmathrm%20dS%3D%5Ciiint_R%5Cnabla%5Ccdot%5Cmathbf%20F%5C%2C%5Cmathrm%20dR)
where
![R](https://tex.z-dn.net/?f=R)
denotes the space bounded by
![S_2](https://tex.z-dn.net/?f=S_2)
. Assuming the vector field is given to be
![\mathbf F(x,y,z)=z^2x\,\mathbf i+(y^3+\tan z)\,\mathbf j+(x^2z+y^2)\,\mathbf k](https://tex.z-dn.net/?f=%5Cmathbf%20F%28x%2Cy%2Cz%29%3Dz%5E2x%5C%2C%5Cmathbf%20i%2B%28y%5E3%2B%5Ctan%20z%29%5C%2C%5Cmathbf%20j%2B%28x%5E2z%2By%5E2%29%5C%2C%5Cmathbf%20k)
then
![\nabla\cdot\mathbf F=\dfrac\partial{\partial x}[z^2x]+\dfrac\partial{\partial y}[y^3+\tan z]+\dfrac\partial{\partial z}[x^2z+y^2]=z^2+3y^2+x^2](https://tex.z-dn.net/?f=%5Cnabla%5Ccdot%5Cmathbf%20F%3D%5Cdfrac%5Cpartial%7B%5Cpartial%20x%7D%5Bz%5E2x%5D%2B%5Cdfrac%5Cpartial%7B%5Cpartial%20y%7D%5By%5E3%2B%5Ctan%20z%5D%2B%5Cdfrac%5Cpartial%7B%5Cpartial%20z%7D%5Bx%5E2z%2By%5E2%5D%3Dz%5E2%2B3y%5E2%2Bx%5E2)
Converting to spherical coordinates, we take
![\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\varphi\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dx%3D%5Crho%5Ccos%5Ctheta%5Csin%5Cvarphi%5C%5Cy%3D%5Crho%5Csin%5Ctheta%5Csin%5Cvarphi%5C%5Cz%3D%5Crho%5Ccos%5Cvarphi%5Cend%7Bcases%7D)
so that the triple integral becomes
![\displaystyle\int_{\varphi=0}^{\varphi=\pi/2}\int_{\theta=0}^{\theta=2\pi}\int_{\rho=0}^{\rho=3}(\rho^2\cos^2\varphi+3\rho^2\sin^2\theta\sin^2\varphi+\rho^2\cos^2\theta\sin^2\varphi)\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cint_%7B%5Cvarphi%3D0%7D%5E%7B%5Cvarphi%3D%5Cpi%2F2%7D%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7B%5Crho%3D0%7D%5E%7B%5Crho%3D3%7D%28%5Crho%5E2%5Ccos%5E2%5Cvarphi%2B3%5Crho%5E2%5Csin%5E2%5Ctheta%5Csin%5E2%5Cvarphi%2B%5Crho%5E2%5Ccos%5E2%5Ctheta%5Csin%5E2%5Cvarphi%29%5Crho%5E2%5Csin%5Cvarphi%5C%2C%5Cmathrm%20d%5Crho%5C%2C%5Cmathrm%20d%5Ctheta%5C%2C%5Cmathrm%20d%5Cvarphi)
![=\displaystyle\int_0^{\pi/2}\int_0^{2\pi}\int_0^3\rho^4(\cos^2\varphi+2\sin^2\theta\sin^2\varphi+\sin^2\varphi)\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E%7B%5Cpi%2F2%7D%5Cint_0%5E%7B2%5Cpi%7D%5Cint_0%5E3%5Crho%5E4%28%5Ccos%5E2%5Cvarphi%2B2%5Csin%5E2%5Ctheta%5Csin%5E2%5Cvarphi%2B%5Csin%5E2%5Cvarphi%29%5C%2C%5Cmathrm%20d%5Crho%5C%2C%5Cmathrm%20d%5Ctheta%5C%2C%5Cmathrm%20d%5Cvarphi)
![=162\pi](https://tex.z-dn.net/?f=%3D162%5Cpi)
Now the integral over
![S](https://tex.z-dn.net/?f=S)
alone will be the difference of the integral over
![S_2](https://tex.z-dn.net/?f=S_2)
and the integral over
![S_1](https://tex.z-dn.net/?f=S_1)
, i.e.
![\displaystyle\iint_{S_2}=\iint_{S_1\cup S}=\iint_{S_1}+\iint_S\implies\iint_S=\iint_{S_2}-\iint_{S_1}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_%7BS_2%7D%3D%5Ciint_%7BS_1%5Ccup%20S%7D%3D%5Ciint_%7BS_1%7D%2B%5Ciint_S%5Cimplies%5Ciint_S%3D%5Ciint_%7BS_2%7D-%5Ciint_%7BS_1%7D)
We can parameterize the points in
![S_1](https://tex.z-dn.net/?f=S_1)
by
![\mathbf s(r,\theta)=\begin{cases}x=r\cos\theta\\y=r\sin\theta\\z=0\end{cases}](https://tex.z-dn.net/?f=%5Cmathbf%20s%28r%2C%5Ctheta%29%3D%5Cbegin%7Bcases%7Dx%3Dr%5Ccos%5Ctheta%5C%5Cy%3Dr%5Csin%5Ctheta%5C%5Cz%3D0%5Cend%7Bcases%7D)
so that the integral over
![S_1](https://tex.z-dn.net/?f=S_1)
is
![\displaystyle\iint_{S_1}\mathbf F\cdot\mathrm dS=\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=3}\mathbf F(x(r,\theta),y(r,\theta),z(r,\theta))\cdot\left(\dfrac{\partial\mathbf s}{\partial r}\times\dfrac{\partial\mathbf s}{\partial\theta}\right)\,\mathrm dr\,\mathrm d\theta](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciint_%7BS_1%7D%5Cmathbf%20F%5Ccdot%5Cmathrm%20dS%3D%5Cint_%7B%5Ctheta%3D0%7D%5E%7B%5Ctheta%3D2%5Cpi%7D%5Cint_%7Br%3D0%7D%5E%7Br%3D3%7D%5Cmathbf%20F%28x%28r%2C%5Ctheta%29%2Cy%28r%2C%5Ctheta%29%2Cz%28r%2C%5Ctheta%29%29%5Ccdot%5Cleft%28%5Cdfrac%7B%5Cpartial%5Cmathbf%20s%7D%7B%5Cpartial%20r%7D%5Ctimes%5Cdfrac%7B%5Cpartial%5Cmathbf%20s%7D%7B%5Cpartial%5Ctheta%7D%5Cright%29%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta)
![=\displaystyle\int_0^{2\pi}\int_0^3(r^3\sin^3\theta\,\mathbf j+r^2\sin^2\theta\,\mathbf k)\cdot(r\,\mathbf k)\,\mathrm dr\,\mathrm d\theta](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E%7B2%5Cpi%7D%5Cint_0%5E3%28r%5E3%5Csin%5E3%5Ctheta%5C%2C%5Cmathbf%20j%2Br%5E2%5Csin%5E2%5Ctheta%5C%2C%5Cmathbf%20k%29%5Ccdot%28r%5C%2C%5Cmathbf%20k%29%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta)
![=\displaystyle\int_0^{2\pi}\int_0^3r^3\sin^2\theta\,\mathrm dr\,\mathrm d\theta](https://tex.z-dn.net/?f=%3D%5Cdisplaystyle%5Cint_0%5E%7B2%5Cpi%7D%5Cint_0%5E3r%5E3%5Csin%5E2%5Ctheta%5C%2C%5Cmathrm%20dr%5C%2C%5Cmathrm%20d%5Ctheta)
![=\dfrac{81\pi}4](https://tex.z-dn.net/?f=%3D%5Cdfrac%7B81%5Cpi%7D4)
So, the integral over
![S](https://tex.z-dn.net/?f=S)
alone evaluates to
Answer:
<u>The standard error of distribution for n = 4 is 5 and for n = 25 is 2.</u>
Step-by-step explanation:
1. Let's review the information given to us for solving the question:
Population mean = 72
Standard deviation = 10
Sample₁ = 4
Sample₂ = 25
2. For finding the standard error of the mean, we use the following formula:
Standard error = Standard deviation / √Size of the sample
Standard error for Sample₁ = 10/√4
<u>Standard error for Sample₁ = 10/2 = 5</u>
Now, let's find the standard error for Sample₂
Standard error for Sample₂ = 10/√25
<u>Standard error for Sample₂ = 10/5 = 2</u>
Fat percentage = Fat weight / Body weight * 100
Assuming that 3/2x^2 is (3/2)x^2
Fat percentage =
Answer:
35 :
t = 6.25 years
(about 6 years 3 months)
Equation:
t = (1/r)(A/P - 1)
Calculation:
First, converting R percent to r a decimal
r = R/100 = 4%/100 = 0.04 per year,
then, solving our equation
t = (1/0.04)((2500/2000) - 1) = 6.25
t = 6.25 years
The time required to get a total amount, principal plus interest, of $2,500.00 from simple interest on a principal of $2,000.00 at an interest rate of 4% per year is 6.25 years (about 6 years 3 months).
36:
The two distances are the same (out and back), so set them equal.
That is done by having a (rate)(time) equal a (rate)(time).
One time is “x” and the other is “4.8-x.”
One rate is 460 and the other is 500.
460 x = 500 (4.8 -x)
460 x = 2400 - 500x
900 x = 2400
x = 2.5 hours for the slower plane.
4.8- x = 2.3 hours for the faster plane.