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

[WMDhamnekar]

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

Relevant Equations

None

But the answer provided is $\frac{15}{4}$ How is that? What is wrong in the above computation of answer?

 

[pasmith]

\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 $$d\boldsymbol{\sigma} = <br /> \begin{pmatrix} 1 \\ 0 \\ \frac{\partial z}{\partial x} \end{pmatrix} \times <br /> \begin{pmatrix} 0 \\ 1 \\ \frac{\partial z}{\partial y} \end{pmatrix}\,dx\,dy<br /> = \begin{pmatrix} - \frac{\partial z}{\partial x} \\ -\frac{\partial z}{\partial y} \\ 1 \end{pmatrix}\,dx\,dy <br /> = \begin{pmatrix} 3 \\ \tfrac32 \\ 1\end{pmatrix}\,dx\,dy$$ and the surface integral then reduces to $$\begin{split}<br /> \iint_\Sigma \mathbf{f} \cdot d\boldsymbol{\sigma} &=<br /> \int_0^1 \int_0^{2(1-x)} 3x^2 + \tfrac32xy + \left(3 - 3x - \tfrac32y\right)^2\,dy \,dx<br /> \\<br /> &= \int_0^1 \int_0^{2(1-x)} \tfrac32 x(2x+y) + \left(3 - \tfrac32(2x+y)\right)^2\,dy\,dx \\<br /> &= \int_0^1 \int_{2x}^2 \tfrac32 xu + 2\left(3 - \tfrac32 u\right)^2\,du\,dx<br /> \end{split}$$ using the substitution $u = 2x + y$ in the inner integral (I think this substitution simplifies the algebra significantly, thereby reducing the scope for errors).

 

[WMDhamnekar]

\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 $$d\boldsymbol{\sigma} =<br /> \begin{pmatrix} 1 \\ 0 \\ \frac{\partial z}{\partial x} \end{pmatrix} \times<br /> \begin{pmatrix} 0 \\ 1 \\ \frac{\partial z}{\partial y} \end{pmatrix}\,dx\,dy<br /> = \begin{pmatrix} - \frac{\partial z}{\partial x} \\ -\frac{\partial z}{\partial y} \\ 1 \end{pmatrix}\,dx\,dy<br /> = \begin{pmatrix} 3 \\ \tfrac32 \\ 1\end{pmatrix}\,dx\,dy$$ and the surface integral then reduces to $$\begin{split}<br /> \iint_\Sigma \mathbf{f} \cdot d\boldsymbol{\sigma} &=<br /> \int_0^1 \int_0^{2(1-x)} 3x^2 + \tfrac32xy + \left(3 - 3x - \tfrac32y\right)^2\,dy \,dx<br /> \\<br /> &= \int_0^1 \int_0^{2(1-x)} \tfrac32 x(2x+y) + \left(3 - \tfrac32(2x+y)\right)^2\,dy\,dx \\<br /> &= \int_0^1 \int_{2x}^2 \tfrac32 xu + 2\left(3 - \tfrac32 u\right)^2\,du\,dx<br /> \end{split}$$ 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.

 

[pasmith]

Z compnent is correct. It is $z\hat{k}$ I wrote it wrongly in the homework statement. Sorry for that.

Then we should have $$\begin{split}<br /> \iint_\Sigma \mathbf{f} \cdot d\boldsymbol{\sigma} &=<br /> \int_0^1 \int_0^{2(1-x)} 3x^2 + \tfrac32xy + \left(3 - 3x - \tfrac32y\right)\,dy \,dx<br /> \\<br /> &= \int_0^1 \int_0^{2(1-x)} \tfrac32 x(2x+y) + \left(3 - \tfrac32(2x+y)\right)\,dy\,dx \\<br /> &= \int_0^1 \int_{2x}^2 \tfrac32 xu + 2\left(3 - \tfrac32 u\right)\,du\,dx<br /> \end{split}$$ 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 $$\int_0^1 \int_{2x}^2 3ux + 3 - \tfrac32u\,du\,dx = \frac 74$$ which is your answer.

 

[Delta2]

@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 $$f(x,y,z)=x^2\hat i+xy\hat j+2z\hat k$$?

 

[pasmith]

@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.

 

[WMDhamnekar]

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.

 

[Delta2]

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...