Is my calc. for triple integral correct? int[B]x^2y^2z^2 dV = 1/5, B={0 <=x<=y<=z<=1}

tsp216

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Is my calc. for triple integral correct? int[B]x^2y^2z^2 dV = 1/5, B={0 <=x<=y<=z<=1}

So I'm asked to calculate the triple integral
\(\displaystyle \iiint_{B} x^2 y^2 z^2 dV\) for a region B defined as 0 <=x<=y<=z<=1.

My setup is therefore

\(\displaystyle \int[0,1]\int[0,z]\int[0,y] x^2 y^2 z^2 dxdydz\). I then went on to evaluate this, and got 1/5. Is this correct?
 
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So I'm asked to calculate the triple integral
\(\displaystyle \iiint_{B} x^2 y^2 z^2 dV\) for a region B defined as 0 <=x<=y<=z<=1.

My setup is therefore

\(\displaystyle \int[0,1]\int[0,z]\int[0,y] x^2 y^2 z^2 dxdydz\). I then went on to evaluate this, and got 1/5. Is this correct?

Off the top of my head, I would expect something not very far away from [(1/2)^2]^3 = 1/64. 1/5 seems a little large.

Please demonstrate any intermediate results. Perhaps your result after the inner-most integral - leaving the other 2 pending?

Note: We use [ t e x ] tags for math text, not the dollar sign tags.
 
Upon recalculation, I got

\(\displaystyle \int_{0}^{1} \int_{0}^{z} \int_{0}^{y} x^2 y^2 z^2 dxdydz\\
=\int_{0}^{1}\int_{0}^{z} \left[ \frac{x^3}{3}\right]_{0}^{y} y^2 z^2 dxdydz\\
=\int_{0}^{1} \int_{0}^{z} \frac{y^3}{3} y^2 z^2 dydz\\
=\int_{0}^{z} \int_{0}^{z} \frac{y^5}{3} z^2 dydz\\
= \int_{0}^{z} \left[ \frac{y^6}{18} \right]_{0}^{z} z^2 dz\\
=\int_{0}^{z} \frac{z^6}{18} z^2 dz\\
=\left[ \frac{z^9}{18\cdot 9} \right]_{0}^{1} = 1/162\)

Does this seem more correct?

Also, if my reply doesn't format properly, could you please send me a link to a tutorial to correctly format it? Thanks.
 
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Thanks to sinx and tkhunny for confirming the answer and for the formatting help.
 
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