Integral xyz When 0<=y<=x<=z<=1??
H Hoseein New member Joined Jun 16, 2021 Messages 1 Jun 16, 2021 #1 Integral xyz When 0<=y<=x<=z<=1??
D Deleted member 4993 Guest Jun 17, 2021 #2 Hoseein said: Integral xyz When 0<=y<=x<=z<=1?? Click to expand... Please post the EXACT problem - preferably a photo if it is in English.
Hoseein said: Integral xyz When 0<=y<=x<=z<=1?? Click to expand... Please post the EXACT problem - preferably a photo if it is in English.
H HallsofIvy Elite Member Joined Jan 27, 2012 Messages 7,763 Jun 17, 2021 #3 So z goes from 0 to 1, for any value of z, x goe from 0 to z, and, for any value of x, y goes from 0 to x. That would be ∫z=01∫x=0z∫y=0xxyzdydxdz\displaystyle \int_{z= 0}^1\int_{x= 0}^z\int_{y=0}^x xyz dydxdz∫z=01∫x=0z∫y=0xxyzdydxdz =∫z=01∫x=0z12[xy2z]y=0xdxdz=12∫z=01∫x=0zx3zdxdz\displaystyle = \int_{z= 0}^1\int_{x= 0}^z \frac{1}{2}\left[xy^2z\right]_{y= 0}^x dxdz= \frac{1}{2}\int_{z= 0}^1\int_{x= 0}^z x^3z dxdz=∫z=01∫x=0z21[xy2z]y=0xdxdz=21∫z=01∫x=0zx3zdxdz =12∫z=0114[x4z]x=0zdz=18∫z=01z5dz\displaystyle = \frac{1}{2}\int_{z= 0}^1 \frac{1}{4}\left[x^4z\right]_{x=0}^z dz= \frac{1}{8}\int_{z= 0}^1 z^5dz=21∫z=0141[x4z]x=0zdz=81∫z=01z5dz =18[16z6]z=01=148\displaystyle = \frac{1}{8}\left[\frac{1}{6}z^6\right]_{z= 0}^1= \frac{1}{48}=81[61z6]z=01=481.
So z goes from 0 to 1, for any value of z, x goe from 0 to z, and, for any value of x, y goes from 0 to x. That would be ∫z=01∫x=0z∫y=0xxyzdydxdz\displaystyle \int_{z= 0}^1\int_{x= 0}^z\int_{y=0}^x xyz dydxdz∫z=01∫x=0z∫y=0xxyzdydxdz =∫z=01∫x=0z12[xy2z]y=0xdxdz=12∫z=01∫x=0zx3zdxdz\displaystyle = \int_{z= 0}^1\int_{x= 0}^z \frac{1}{2}\left[xy^2z\right]_{y= 0}^x dxdz= \frac{1}{2}\int_{z= 0}^1\int_{x= 0}^z x^3z dxdz=∫z=01∫x=0z21[xy2z]y=0xdxdz=21∫z=01∫x=0zx3zdxdz =12∫z=0114[x4z]x=0zdz=18∫z=01z5dz\displaystyle = \frac{1}{2}\int_{z= 0}^1 \frac{1}{4}\left[x^4z\right]_{x=0}^z dz= \frac{1}{8}\int_{z= 0}^1 z^5dz=21∫z=0141[x4z]x=0zdz=81∫z=01z5dz =18[16z6]z=01=148\displaystyle = \frac{1}{8}\left[\frac{1}{6}z^6\right]_{z= 0}^1= \frac{1}{48}=81[61z6]z=01=481.