Answer: 16 pi or ≈50.27
Explanation: Formula for a cylinder:
V = pi r ^2 h
V = pi (2)^2 4
V = 4 pi (4)
V = 16pi
Answer:
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Step-by-step explanation:
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Answer:
(x + 6 )(x + 6 )
2 numbers to add to 12 but also multiply to get 36
Consider the top half of a sphere centered at the origin with radius
![r](https://tex.z-dn.net/?f=r)
, which can be described by the equation
![z=\sqrt{r^2-x^2-y^2}](https://tex.z-dn.net/?f=z%3D%5Csqrt%7Br%5E2-x%5E2-y%5E2%7D)
and consider a plane
![z=h](https://tex.z-dn.net/?f=z%3Dh)
with
![0](https://tex.z-dn.net/?f=0%3Ch%3Cr)
. Call the region between the two surfaces
![R](https://tex.z-dn.net/?f=R)
. The volume of
![R](https://tex.z-dn.net/?f=R)
is given by the triple integral
![\displaystyle\iiint_R\mathrm dV=\int_{-\sqrt{r^2-h^2}}^{\sqrt{r^2-h^2}}\int_{-\sqrt{r^2-h^2-x^2}}^{\sqrt{r^2-h^2-x^2}}\int_h^{\sqrt{r^2-x^2-y^2}}\mathrm dz\,\mathrm dy\,\mathrm dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciiint_R%5Cmathrm%20dV%3D%5Cint_%7B-%5Csqrt%7Br%5E2-h%5E2%7D%7D%5E%7B%5Csqrt%7Br%5E2-h%5E2%7D%7D%5Cint_%7B-%5Csqrt%7Br%5E2-h%5E2-x%5E2%7D%7D%5E%7B%5Csqrt%7Br%5E2-h%5E2-x%5E2%7D%7D%5Cint_h%5E%7B%5Csqrt%7Br%5E2-x%5E2-y%5E2%7D%7D%5Cmathrm%20dz%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20dx)
Converting to polar coordinates will help make this computation easier. Set
![\begin{cases}x=\rho\cos\theta\sin\varphi\\y=\rho\sin\theta\sin\varphi\\z=\rho\cos\var\phi\end{cases}\implies\mathrm dx\,\mathrm dy\,\mathrm dz=\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi](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%5Cvar%5Cphi%5Cend%7Bcases%7D%5Cimplies%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20dz%3D%5Crho%5E2%5Csin%5Cvarphi%5C%2C%5Cmathrm%20d%5Crho%5C%2C%5Cmathrm%20d%5Ctheta%5C%2C%5Cmathrm%20d%5Cvarphi)
Now, the volume can be computed with the integral
![\displaystyle\iiint_R\mathrm dV=\int_0^{2\pi}\int_0^{\arctan\frac{\sqrt{r^2-h^2}}h}\int_{h\sec\varphi}^r\rho^2\sin\varphi\,\mathrm d\rho\,\mathrm d\varphi\,\mathrm d\theta](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciiint_R%5Cmathrm%20dV%3D%5Cint_0%5E%7B2%5Cpi%7D%5Cint_0%5E%7B%5Carctan%5Cfrac%7B%5Csqrt%7Br%5E2-h%5E2%7D%7Dh%7D%5Cint_%7Bh%5Csec%5Cvarphi%7D%5Er%5Crho%5E2%5Csin%5Cvarphi%5C%2C%5Cmathrm%20d%5Crho%5C%2C%5Cmathrm%20d%5Cvarphi%5C%2C%5Cmathrm%20d%5Ctheta)
You should get
Since Joann receives $10 per hour for the first 35 hours, so she receives $350 (35 times $10) before she starts receiving extra hours.
Since she received $515, we know that she worked extra hours.
Let's use the variable x to represent the number of extra hours worked.
So we can write the following equation for the total payment received:
![\begin{gathered} 10\cdot35+15\cdot x=515 \\ 350+15x=515 \\ 15x=515-350 \\ 15x=165 \\ x=11 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%2010%5Ccdot35%2B15%5Ccdot%20x%3D515%20%5C%5C%20350%2B15x%3D515%20%5C%5C%2015x%3D515-350%20%5C%5C%2015x%3D165%20%5C%5C%20x%3D11%20%5Cend%7Bgathered%7D)
She worked 11 extra hours. Adding this to the 35 regular hours, so she worked for 46 hours.