Answer:
M = 5742π
Step-by-step explanation:
Given:-
- Find the mass of a solid with the density ( ρ ):
ρ ( r, θ , z ) = 1 + z / 81
- The solid is bounded by the planes:
0 ≤ z ≤ 81 - r^2
0 ≤ r ≤ 9
Find:-
Find the mass of the solid paraboloid
Solution:-
- The mass (M) of any solid body is given by the following triple integral formulation:
- We can write the above expression in cylindrical coordinates:
![M = \int\limits\int\limits_r\int\limits_z {r*p(r,theta,z)} \, dz.dr.dtheta \\\\M = \int\limits\int\limits_r\int\limits_z {r*[ 1 + \frac{z}{81}] } \, dz.dr.dtheta\\\\](https://tex.z-dn.net/?f=M%20%3D%20%5Cint%5Climits%5Cint%5Climits_r%5Cint%5Climits_z%20%7Br%2Ap%28r%2Ctheta%2Cz%29%7D%20%5C%2C%20dz.dr.dtheta%20%5C%5C%5C%5CM%20%3D%20%5Cint%5Climits%5Cint%5Climits_r%5Cint%5Climits_z%20%7Br%2A%5B%201%20%2B%20%5Cfrac%7Bz%7D%7B81%7D%5D%20%7D%20%5C%2C%20dz.dr.dtheta%5C%5C%5C%5C)
- Perform integration:
![M = \int\limits\int\limits_r{r*[ z + \frac{z^2}{162}] } \,|_0^8^1^-^r^2 dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{(81-r^2)^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + \frac{6561 -162r + r^2}{162}] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{r*[ 81-r^2 + 40.5 -r +\frac{r^2}{162} ] } \, dr.dtheta\\\\M = \int\limits\int\limits_r{[ 121.5r-r^2 -\frac{161r^3}{162} ] } \, dr.dtheta\\\\](https://tex.z-dn.net/?f=M%20%3D%20%5Cint%5Climits%5Cint%5Climits_r%7Br%2A%5B%20z%20%2B%20%5Cfrac%7Bz%5E2%7D%7B162%7D%5D%20%7D%20%5C%2C%7C_0%5E8%5E1%5E-%5Er%5E2%20dr.dtheta%5C%5C%5C%5CM%20%3D%20%5Cint%5Climits%5Cint%5Climits_r%7Br%2A%5B%2081-r%5E2%20%2B%20%5Cfrac%7B%2881-r%5E2%29%5E2%7D%7B162%7D%5D%20%7D%20%5C%2C%20dr.dtheta%5C%5C%5C%5CM%20%3D%20%5Cint%5Climits%5Cint%5Climits_r%7Br%2A%5B%2081-r%5E2%20%2B%20%5Cfrac%7B6561%20-162r%20%2B%20r%5E2%7D%7B162%7D%5D%20%7D%20%5C%2C%20dr.dtheta%5C%5C%5C%5CM%20%3D%20%5Cint%5Climits%5Cint%5Climits_r%7Br%2A%5B%2081-r%5E2%20%2B%2040.5%20-r%20%2B%5Cfrac%7Br%5E2%7D%7B162%7D%20%5D%20%7D%20%5C%2C%20dr.dtheta%5C%5C%5C%5CM%20%3D%20%5Cint%5Climits%5Cint%5Climits_r%7B%5B%20121.5r-r%5E2%20-%5Cfrac%7B161r%5E3%7D%7B162%7D%20%5D%20%7D%20%5C%2C%20dr.dtheta%5C%5C%5C%5C)
![M = 2*\int\limits_0^\pi {[ 121.5r^2-r^3 -\frac{161r^4}{162} ] } |_0^6 \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 121.5(6)^2-(6)^3 -\frac{161(6)^4}{162} ] } \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 4375-216 -1288] } \, dtheta\\\\M = 2*\int\limits_0^\pi {[ 2871] } \, dtheta\\\\M = 5742\pi kg](https://tex.z-dn.net/?f=M%20%3D%202%2A%5Cint%5Climits_0%5E%5Cpi%20%7B%5B%20121.5r%5E2-r%5E3%20-%5Cfrac%7B161r%5E4%7D%7B162%7D%20%5D%20%7D%20%7C_0%5E6%20%5C%2C%20dtheta%5C%5C%5C%5CM%20%3D%202%2A%5Cint%5Climits_0%5E%5Cpi%20%7B%5B%20121.5%286%29%5E2-%286%29%5E3%20-%5Cfrac%7B161%286%29%5E4%7D%7B162%7D%20%5D%20%7D%20%20%5C%2C%20dtheta%5C%5C%5C%5CM%20%3D%202%2A%5Cint%5Climits_0%5E%5Cpi%20%7B%5B%204375-216%20-1288%5D%20%7D%20%20%5C%2C%20dtheta%5C%5C%5C%5CM%20%3D%202%2A%5Cint%5Climits_0%5E%5Cpi%20%7B%5B%202871%5D%20%7D%20%20%5C%2C%20dtheta%5C%5C%5C%5CM%20%3D%205742%5Cpi%20%20kg)
- The mass evaluated is M = 5742π
Answer:
4 and me too
Step-by-step explanation:
I think it would take about 1 hour and 17 minutes
Answer:
$4,400
Step-by-step explanation:
Key: B stands for balance.
b+308=4708
b=4,708-308
b=$4,400
Answer:
The width of the rectangle is 14'8", and the length is 33'4"
Step-by-step explanation:
We're given two pieces of information:
The length is eight more than twice the width:
The perimeter is 96 feet:
We also need to apply one more piece of information that is not provided here, and that is the relationship between the perimeter of a rectangle, and it's length and width:
We can solve for w by plugging the other two values into the last:
Now we can find the length by plugging w into the first equation:
One third of a foot is four inches, so the width is 14'8" and the length is 33'4"
To make sure our answer is correct, we should plug those numbers back into the area equation and see if we're right:
So we know our answer's correct
