Evaluating $\iint\limits_{\sum} f \cdot d\sigma $ on Cube S

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In summary, we can evaluate the surface integral of $f(x,y,z) = xi + yj + zk$ over the boundary of the solid cube $S= \{(x,y,z) = 0\leq x,y,z \leq 1)\}$ without using the Divergence Theorem by parameterizing the six faces of the cube and using the formula for a surface integral of a vector field. The result is 3, which is the sum of the integrals over each individual face.
  • #1
WMDhamnekar
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Without using the Divergence Theorem Evaluate the surface integral $\iint\limits_{\sum} f \cdot d\sigma $ of $f(x,y,z) = xi+ yj + zk , \sum: $ boundary of the solid cube S= $\{(x,y,z) = 0\leq x,y,z \leq 1)\}$

My attempt:
Here we have to use the following definition of surface integral.
1657809382885.png


Note that there will be a different outward unit normal vector to each of the six faces of the cube.
 
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Hi Dhamnekar Winod,

The formula you posted is for a real-valued function, but the function $f$ you provided is a vector field. You will need to use the formula for a surface integral of a vector field: $$\iint\limits_{R}f(x(u,v), y(u,v), z(u,v))\cdot \left(\frac{\partial\mathbf{r}}{\partial u}\times \frac{\partial\mathbf{r}}{\partial v} \right)du\,dv \qquad(*)$$ See the Wikipedia - Surface Integrals of Vector Fields.

After that, think about how to parameterize each of the six faces of $\Sigma$. For example, the "bottom" face can be parameterized as $x(u,v) = u$, $y(u,v) = v$, and $z(u,v) = 0$, where $0\leq u, v\leq 1$; i.e., $\mathbf{r}(u,v)= u\mathbf{i} + v\mathbf{j} + 0\mathbf{k}$ for $0\leq u, v\leq 1$. Using this parameterization, calculate the integral in $(*)$. Note: You will need to use the negative of the cross product $\frac{\partial\mathbf{r}}{\partial u}\times \frac{\partial\mathbf{r}}{\partial v}$ to ensure you get an outward pointing normal vector to the "bottom" of $\Sigma$. Then repeat the process for each of the other 5 faces of $\Sigma$.
 
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  • #3
This integral will be over the 6 faces of the cube. Do each separately. One face is z= 0 for x and y each from 0 to 1. On z= 0, the vector field zk= 0k so the integral is 0. That is also true on the faces x= 0 and y= 0. On z= 1 we integrate zk= 1k so we are integrating 1 for both x and y from 0 to 1. That is 1 times the area of the square, which is also 1, so the integral is 1. This is also true on the faces x= 1 and y= 1 so the entire surface integral is 3.
 
  • #4
HallsofIvy said:
This integral will be over the 6 faces of the cube. Do each separately. One face is z= 0 for x and y each from 0 to 1. On z= 0, the vector field zk= 0k so the integral is 0. That is also true on the faces x= 0 and y= 0. On z= 1 we integrate zk= 1k so we are integrating 1 for both x and y from 0 to 1. That is 1 times the area of the square, which is also 1, so the integral is 1. This is also true on the faces x= 1 and y= 1 so the entire surface integral is 3.
f(x,y,z) = xi + yj + zk , $ \vert \frac{dr}{du} \times \frac{dr}{dv} \vert =1$ So, $ \iint\limits_{\sum} = f\cdot d\sigma =\displaystyle\int_0^1 \displaystyle\int_0^{1-x} (x+y+1-(x+y))\cdot (1)dydx=\frac12$

This surface integral is for one face of the cube. So entire surface integral will be $\frac12 \times 6 =3 $
 
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1. What is the purpose of evaluating $\iint\limits_{\sum} f \cdot d\sigma $ on a cube?

The purpose of evaluating $\iint\limits_{\sum} f \cdot d\sigma $ on a cube is to calculate the double integral of a function over a surface, which represents the volume under the function in three-dimensional space.

2. How is the cube defined in this context?

In this context, the cube is defined as a three-dimensional shape with six square faces, each of which has equal side lengths. It is often represented by the Cartesian coordinates (x,y,z) and is a common shape used in mathematics and physics.

3. What does $\sum$ represent in the double integral?

In the double integral, $\sum$ represents the surface over which the function is being integrated. In this case, it is the cube, but it can also be any other surface in three-dimensional space.

4. How is the value of the double integral calculated on a cube?

The value of the double integral on a cube is calculated by dividing the cube into smaller, equal-sized rectangles and approximating the volume under the function by summing the areas of these rectangles. As the number of rectangles increases, the approximation becomes more accurate.

5. What factors can affect the accuracy of the double integral on a cube?

The accuracy of the double integral on a cube can be affected by the number of rectangles used in the approximation, the complexity of the function being integrated, and the size of the cube. Using a larger number of smaller rectangles and a smaller cube can lead to a more accurate result.

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