Help Integrating Tripple Integral for x+y+z=1

  • Thread starter Thread starter Alem2000
  • Start date Start date
  • Tags Tags
    Integrating
AI Thread Summary
The discussion revolves around evaluating the triple integral of z over a tetrahedron defined by the planes x=0, y=0, z=0, and x+y+z=1. The user has set up the integral correctly but struggles with the integration process. They have provided their integral setup and requested assistance with the final steps. A suggestion was made to use the y-anti-derivative for the inner integral, specifically -1/3(1-x-y)³. The conversation emphasizes the need for clarity in the integration process to complete the evaluation.
Alem2000
Messages
117
Reaction score
0
the question is "evaluate \iiint z \,dv, of a solid tetrahedron bounded by the four planes x=0,y=0,z=0, and x+y+z=1"
I can set up the problem correctly but i can't seem to integrate it right
\int_{0}^1 \int_{0}^{1-x} \int_{0}^{1-x-y} z dzdydx
(1/2) \int_{0}^1 \int_{0}^{1-x} (1-x-y)^2dydx
can somone please show me the last few steps of this problem? :confused:
 
Physics news on Phys.org
Open the brackets and integrate each term with corresponding limits...

Daniel.
 
For the evaluation of the inner integral, use the y-anti-derivative :
-\frac{1}{3}(1-x-y)^{3}
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top