# Having Difficuilty With Evaluating A Triple Integral

Here's the question $$\int_{2}^{4}\int_{1}^{2}\int_{0}^{4}xy(z+2)dxdydz$$

Is it right to distribute $xy$ to $z$ and $2$ obtaining $$\int_{2}^{4}\int_{1}^{2}\int_{0}^{4}(xyz+2xy)dxdydz ?$$

and then start with the inner most integral i.e $$\int_{0}^{4}(xyz + 2xy)dx$$ and integrate only with respect to $x$ keeping the rest constant and evaluating between $0$ and $4$. $$i.e \left (\frac{x^2}{2}yz + x^2y \right )_{0}^{4}$$ $$=8yz + 16y$$

Does the next step invlove integrating the above expression $w.r.t$ $y$ keeping $z$ constant?i.e$$\int_{1}^{2}(8yz + 16y)dy$$

If i follow the above step i get the final answer to be $180$ which is obviously wrong.As per the text book that i'am using it's $120$.

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Yes you're doing it right. Just evaluate each integral seperately. You have probably done some integration mistake in the second or third integral. Here's my solution: $$\int_0^4 (xyz+2xy)\,\mathrm dx=\left[\frac{1}{2}yzx^2+yx^2\right]_0^4=8yz+16y$$ and $$\int_1^2(8yz+16y)\,\mathrm dy=\left[4zy^2+8y^2\right]_1^2=16z+32-4z-8=12z+24$$ and $$\int_2^4(12z+24)\,\mathrm dz=\left[6z^2+24z\right]_2^4=96+96-24-48=120.$$