Question 27.16: Evaluate I = ∫x=0 ^x=1 ∫y=0 ^y=1 ∫z=0 ^z=1 x + y + z dz dy ......
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} xStep-by-Step
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} xwhere, as before, the inner integral is performed first, integrating with respect to z, with x and y being treated as constants. So
\begin{aligned} 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} \mathrm{~d} x \\ & =\left[\frac{x^2}{2}+x\right]_0^1 \\ & =\frac{3}{2} \end{aligned}Consideration of the limits of integration shows that the integral is evaluated over a unit cube.
Related Answered Questions
Question: 27.19
Verified Answer:
(a) The open face is highlighted in Figure 27.26. ...
Question: 27.18
Verified Answer:
So, at any point P with coordinates (x, y, z), the...
Question: 27.8
Verified Answer:
If \mathbf{v} is derivable from the...
Question: 27.7
Verified Answer:
We find curl F:
\begin{aligned} \nabla \tim...
Question: 27.5
Verified Answer:
\int_C \mathbf{F}(x, y) \cdot \mathrm{d} \m...
Question: 27.4
Verified Answer:
We evaluate this integral in two parts because the...
Question: 27.3
Verified Answer:
We compare the integrand with the standard form an...
Question: 27.1
Verified Answer:
The work done by gravity is found by evaluating th...
Question: 27.6
Verified Answer:
(a) When t = 1, x = A, y = 1 and z = 6, and so P h...
Question: 27.26
Verified Answer:
(a) The open face is highlighted in Figure 27.26. ...