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I am having problem in integrating the equation below. If I integrate it w.r.t x then w.r.t y and then w.r.t z, the answers comes out to be 0 but the actual answer is 52. Please help out. Thanks enter image description here

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Start with integrating w.r.t. $x$ and treat $y,z$ as constants, $ \int C +Ax^2 dx $. Then continue... –  draks ... Jun 27 '12 at 13:46

1 Answer 1

$$\begin{align} \int_1^3 (6yz^3+6x^2y)\,dx &= \left[6xyz^3+2x^3y\right]_{x=1}^3 =12yz^3+52y \\ \int_0^1 (12yz^3+52y)\,dy &=\left[6y^2z^3+26y^2\right]_{y=0}^1 =6z^3+26 \\ \int_{-1}^1 (6z^3+26)\,dz &= \left[\frac{3}{2}z^4+26z\right]_{z=-1}^1 = 52 . \end{align}$$

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