Evaluate the surface integral ##\iint\limits_{\sum} f\cdot d\sigma##

WMDhamnekar
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Homework Statement
Evaluate the surface integral ##\iint\limits_{\sum}f \cdot d\sigma## where ##f(x,y,z) = x^2\hat{i} + xy\hat{j} + z^2\hat{k}## and ##\sum## is the part of the plane 6x + 3y + 2z =6 with ##x \geq 0, y \geq 0, ## and ##z \geq 0 ## with the outward unit normal n pointing in the positive z direction
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But the answer provided is ##\frac{15}{4} ## How is that? What is wrong in the above computation of answer?
 

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\sum produces the summation operator \sum; \Sigma produces a capital sigma \Sigma.

You haven't used the correct z component of \mathbf{f}; it should be z^2, not z as you have.

As regards the rest of your working, it is simplest to just use x and y as parameters rather than introducing u and v. Then d\boldsymbol{\sigma} = <br /> \begin{pmatrix} 1 \\ 0 \\ \frac{\partial z}{\partial x} \end{pmatrix} \times <br /> \begin{pmatrix} 0 \\ 1 \\ \frac{\partial z}{\partial y} \end{pmatrix}\,dx\,dy<br /> = \begin{pmatrix} - \frac{\partial z}{\partial x} \\ -\frac{\partial z}{\partial y} \\ 1 \end{pmatrix}\,dx\,dy <br /> = \begin{pmatrix} 3 \\ \tfrac32 \\ 1\end{pmatrix}\,dx\,dy<br /> and the surface integral then reduces to <br /> \begin{split}<br /> \iint_\Sigma \mathbf{f} \cdot d\boldsymbol{\sigma} &amp;=<br /> \int_0^1 \int_0^{2(1-x)} 3x^2 + \tfrac32xy + \left(3 - 3x - \tfrac32y\right)^2\,dy \,dx<br /> \\<br /> &amp;= \int_0^1 \int_0^{2(1-x)} \tfrac32 x(2x+y) + \left(3 - \tfrac32(2x+y)\right)^2\,dy\,dx \\<br /> &amp;= \int_0^1 \int_{2x}^2 \tfrac32 xu + 2\left(3 - \tfrac32 u\right)^2\,du\,dx<br /> \end{split}<br /> using the substitution u = 2x + y in the inner integral (I think this substitution simplifies the algebra significantly, thereby reducing the scope for errors).
 
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pasmith said:
\sum produces the summation operator \sum; \Sigma produces a capital sigma \Sigma.

You haven't used the correct z component of \mathbf{f}; it should be z^2, not z as you have.

As regards the rest of your working, it is simplest to just use x and y as parameters rather than introducing u and v. Then d\boldsymbol{\sigma} =<br /> \begin{pmatrix} 1 \\ 0 \\ \frac{\partial z}{\partial x} \end{pmatrix} \times<br /> \begin{pmatrix} 0 \\ 1 \\ \frac{\partial z}{\partial y} \end{pmatrix}\,dx\,dy<br /> = \begin{pmatrix} - \frac{\partial z}{\partial x} \\ -\frac{\partial z}{\partial y} \\ 1 \end{pmatrix}\,dx\,dy<br /> = \begin{pmatrix} 3 \\ \tfrac32 \\ 1\end{pmatrix}\,dx\,dy<br /> and the surface integral then reduces to <br /> \begin{split}<br /> \iint_\Sigma \mathbf{f} \cdot d\boldsymbol{\sigma} &amp;=<br /> \int_0^1 \int_0^{2(1-x)} 3x^2 + \tfrac32xy + \left(3 - 3x - \tfrac32y\right)^2\,dy \,dx<br /> \\<br /> &amp;= \int_0^1 \int_0^{2(1-x)} \tfrac32 x(2x+y) + \left(3 - \tfrac32(2x+y)\right)^2\,dy\,dx \\<br /> &amp;= \int_0^1 \int_{2x}^2 \tfrac32 xu + 2\left(3 - \tfrac32 u\right)^2\,du\,dx<br /> \end{split}<br /> using the substitution u = 2x + y in the inner integral (I think this substitution simplifies the algebra significantly, thereby reducing the scope for errors).
Z compnent is correct. It is ##z\hat{k}## I wrote it wrongly in the homework statement. Sorry for that.
 
WMDhamnekar said:
Z compnent is correct. It is ##z\hat{k}## I wrote it wrongly in the homework statement. Sorry for that.

Then we should have <br /> \begin{split}<br /> \iint_\Sigma \mathbf{f} \cdot d\boldsymbol{\sigma} &amp;=<br /> \int_0^1 \int_0^{2(1-x)} 3x^2 + \tfrac32xy + \left(3 - 3x - \tfrac32y\right)\,dy \,dx<br /> \\<br /> &amp;= \int_0^1 \int_0^{2(1-x)} \tfrac32 x(2x+y) + \left(3 - \tfrac32(2x+y)\right)\,dy\,dx \\<br /> &amp;= \int_0^1 \int_{2x}^2 \tfrac32 xu + 2\left(3 - \tfrac32 u\right)\,du\,dx<br /> \end{split}<br /> which does give the result \frac{15}4. However, looking at it again I cnanot justify the factor of 2 which mysteriously appears in front of the second term in the final line, and removing it I then get <br /> \int_0^1 \int_{2x}^2 3ux + 3 - \tfrac32u\,du\,dx = \frac 74 which is your answer.
 
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@pasmith e hehe what do you mean you cannot justify that factor of 2...So the correct result is 7/4 afterall?

Ah maybe I see now, maybe you mean that that z^2 is meant to be 2z instead?

@WMDhamnekar can you please make sure which exactly function is f? Is it $$f(x,y,z)=x^2\hat i+xy\hat j+2z\hat k$$?
 
Delta2 said:
@pasmith e hehe what do you mean you cannot justify that factor of 2...So the correct result is 7/4 afterall?

Between the second and third lines the integrand somehow changes from 3x^2 + 3xy/2 + z to 3x^2 + 3xy/2 + 2z which I think is a typo on my part. The fact that this yields what the OP says is the given answer may or may not be coincidence. But without that erroneous factor of 2 I obtain 7/4. So either the given answer is incorrect or the OP has transcribed the problem incorrectly.
 
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pasmith said:
Between the second and third lines the integrand somehow changes from 3x^2 + 3xy/2 + z to 3x^2 + 3xy/2 + 2z which I think is a typo on my part. The fact that this yields what the OP says is the given answer may or may not be coincidence. But without that erroneous factor of 2 I obtain 7/4. So either the given answer is incorrect or the OP has transcribed the problem incorrectly.
If f(x,y,z) = x2i + xyj + 3zk , then we get the answer ##\frac{15}{4}##. So, there may be typographical error in the book.
 
WMDhamnekar said:
If f(x,y,z) = x2i + xyj + 3zk , then we get the answer ##\frac{15}{4}##. So, there may be typographical error in the book.
you mean ##+2z\hat k##... Seems that third term is cursed in this thread lol...
 

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