Bounds of triple integral

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=01x=0zy=0xxyzdydxdz\displaystyle \int_{z= 0}^1\int_{x= 0}^z\int_{y=0}^x xyz dydxdz

=z=01x=0z12[xy2z]y=0xdxdz=12z=01x=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
=12z=0114[x4z]x=0zdz=18z=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
=18[16z6]z=01=148\displaystyle = \frac{1}{8}\left[\frac{1}{6}z^6\right]_{z= 0}^1= \frac{1}{48}.
 
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