Question 27.14: Evaluate I = ∫x=0ˆx=1 ∫y=0ˆy=1 ∫z=0ˆz=1 x+y+zdzdydx...

Engineering Mathematics A foundation for Electronic, Electrical, Communications and Systems Engineers [1040491]

Evaluate

I=\int_{x=0}^{x=1}\int_{y=0}^{y=1}\int_{z=0}^{z=1}x+y+z\,\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}x

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What is meant by this expression is

$I=\int_{x=0}^{x=1}\left(\int_{y=0}^{y=1}\left(\int_{z=0}^{z=1}x+y+z\,\mathrm{d}z\right)\,\mathrm{d}y\right)\,\mathrm{d}x$ where, as before, the inner integral is performed first, integrating with respect to z, with x and y being treated as constants. So

I=\int_{x=0}^{x=1}\left(\int_{y=0}^{y=1}\left[x z+y z+{\frac{z^{2}}{2}}\right]_{0}^{1}\mathrm{d}y\right)\,\mathrm{d}x\\

=\int_{x=0}^{x=1}\left(\int_{y=0}^{y=1}x+y+{\frac{1}{2}}\,\mathrm{d}y\right)\,\mathrm{d}x\\

=\int_{x=0}^{x=1}\left[x y+{\frac{y^{2}}{2}}+{\frac{1}{2}}y\right]_{0}^{1}\mathrm{d}x\\

= \int_{x=0}^{x=1} x+ \frac{1}{2} + \frac{1}{2} dx \\

=\left[{\frac{x^{2}}{2}}+x\right]_{0}^{1}\\

={\frac{3}{2}}

Consideration of the limits of integration shows that the integral is evaluated over a unit cube.

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