Parameterize the ellipsoid using the augmented spherical coordinates:
![\begin{cases}x=\frac18\rho\cos\theta\sin\varphi\\\\y=\frac18\rho\sin\theta\sin\varphi\\\\z=\frac38\rho\cos\varphi\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dx%3D%5Cfrac18%5Crho%5Ccos%5Ctheta%5Csin%5Cvarphi%5C%5C%5C%5Cy%3D%5Cfrac18%5Crho%5Csin%5Ctheta%5Csin%5Cvarphi%5C%5C%5C%5Cz%3D%5Cfrac38%5Crho%5Ccos%5Cvarphi%5Cend%7Bcases%7D)
Then the Jacobian for the change of coordinates is
![\mathbf J=\dfrac{\partial(x,y,z)}{\partial(\rho,\theta,\varphi)}=\begin{bmatrix}\frac18\cos\theta\sin\varphi&-\frac18\rho\sin\theta\sin\varphi&\frac18\rho\cos\theta\cos\varphi\\\\\frac18\sin\theta\sin\varphi&\frac18\rho\cos\theta\sin\varphi&\frac18\rho\sin\theta\cos\varphi\\\frac38\cos\varphi&0&-\frac38\rho\sin\varphi\end{bmatrix}](https://tex.z-dn.net/?f=%5Cmathbf%20J%3D%5Cdfrac%7B%5Cpartial%28x%2Cy%2Cz%29%7D%7B%5Cpartial%28%5Crho%2C%5Ctheta%2C%5Cvarphi%29%7D%3D%5Cbegin%7Bbmatrix%7D%5Cfrac18%5Ccos%5Ctheta%5Csin%5Cvarphi%26-%5Cfrac18%5Crho%5Csin%5Ctheta%5Csin%5Cvarphi%26%5Cfrac18%5Crho%5Ccos%5Ctheta%5Ccos%5Cvarphi%5C%5C%5C%5C%5Cfrac18%5Csin%5Ctheta%5Csin%5Cvarphi%26%5Cfrac18%5Crho%5Ccos%5Ctheta%5Csin%5Cvarphi%26%5Cfrac18%5Crho%5Csin%5Ctheta%5Ccos%5Cvarphi%5C%5C%5Cfrac38%5Ccos%5Cvarphi%260%26-%5Cfrac38%5Crho%5Csin%5Cvarphi%5Cend%7Bbmatrix%7D)
which has determinant
![\det\mathbf J=-\dfrac3{512}\rho^2\sin\varphi](https://tex.z-dn.net/?f=%5Cdet%5Cmathbf%20J%3D-%5Cdfrac3%7B512%7D%5Crho%5E2%5Csin%5Cvarphi)
Then the volume of the ellipsoid is given by
![\displaystyle\iiint_E\mathrm dx\,\mathrm dy\,\mathrm dz=\iiint_E|\det\mathbf J|\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\varphi](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ciiint_E%5Cmathrm%20dx%5C%2C%5Cmathrm%20dy%5C%2C%5Cmathrm%20dz%3D%5Ciiint_E%7C%5Cdet%5Cmathbf%20J%7C%5C%2C%5Cmathrm%20d%5Crho%5C%2C%5Cmathrm%20d%5Ctheta%5C%2C%5Cmathrm%20d%5Cvarphi)
where
![E](https://tex.z-dn.net/?f=E)
denotes the spaced contained by the ellipsoid. In particular, we have the definite integral and volume
What I should do as you did to someone else is “Too bad XDDDD” but because I’m nice, I’m going to help you out, so convert the fraction into an improper fraction meaning 2+5/9 becomes 23/9, now times that by 3 and you will get 23/3, converting the rest of the answers into improper fractions will help you find the answer, which is 7+6/9 or choice 1
There are 8 marbles so we want 8c2 which is 8*7/2 = 28
Answer:
The shadow length is 67
Step-by-step explanation:
solution shown in the picture
Answer:
Therefore Area of Rectangle IJKL is 108 unit².
Step-by-step explanation:
Given:
The vertices of a Rectangle are
I ( 6 , 8)
J ( 6 , -1)
K ( -6 , -1)
L ( -6 ,8)
To Find:
Area of Rectangle IJKL= ?
Solution:
First we will find the Length and Width of Rectangle by Distance Formula ,
Substituting the values we get
Similarly for JK we will have,
Now
Now Area of Rectangle is given by
Substituting the values we get
Therefore Area of Rectangle IJKL is 108 unit².