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
The closest option to the actual answer is a.
1 5/18 is the actual answer.
If you make both the denominators the same, then you can actually add the fractions together.
To make both of the denominators the same, you need to multiply 5/6 by 9 and 4/9 by 6, which would result in 45/54 + 24/54= 69/54 = 23/18. If we convert it to a mixed fraction, it would result into 1 5/18.
Answer:
the answer is B
Step-by-step explanation:
^^
Answer:
34000cm or 340m
Step-by-step explanation:
We have to find total surface area.
dimensions of 1 wooden box =
l = 1m = 100cm
b = 80cm
h = 25cm
∴ In the case of 2 =
l = 100cm
b= 80cm
h = 25 + 25 = 50cm
T.S.A = 2(lb + lh + bh)
= 2(100*80 + 100*50+ 80*50)
= 2(8000+5000+4000)
= 2*17000
= 34000cm
= 340m
The wording of the question kinda doesn't make sense in some parts, but I'll try anyway. 1 plus 4 is 5, and since BC is 5, x is 1 for BC. -3 plus 14 is 11, so x is -3 for AB.
