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Given $$\int_{\gamma}(x+y)dS$$ where $\gamma$ is defined as triangle connecting points $(0,0)$, $(0,1)$, $(1,0)$

I'm having some issues transforming this integral to normal Riemann form. My try:
$$\int_{\gamma}(x+y)dS=\int_{0}^{1}(x+0)\sqrt{1+0}dx+\int_{0}^{1}(0+y)\sqrt{0+1}dy+{\color{Red} {2\int_{0}^{0.5}(x-x)dx} } = 1 + {\color{Red} 0 }$$ I marked red the incorrect one (incorrect for me). Maybe a stupid question, but how should I transform it to not get zero? And I'm not looking for geometrical solution, I'm new in line integrals and still want to learn the concept behind it.

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  • $\begingroup$ I find your picture confusing: your line integral should be in the $xy$ plane, what's going on in 3 dimensions? $\endgroup$ – onamoonlessnight Nov 11 '15 at 20:17
  • $\begingroup$ I think you're right, this picture is wrong, I'll better remove it. $\endgroup$ – shcolf Nov 11 '15 at 20:22
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A note on notation: it is good taste to write line integrals as $\int f dr$, and reserve the notation $\int f dS$ for surface integrals.

To compute the integral $$\int_{\gamma} (x + y) dr$$ along the triangle with vertices at $(0,0)$, $(0,1)$, $(1,0)$, first draw a picture of $\gamma$:enter image description here

If orientation is not explicitly stated, it is common to assume that the integral goes anticlockwise. The direction of the integral of course matters only up to a sign. Split your integral into the three lines that make up your triangle, $$\int_{\gamma} (x+y) dr = \int_{\rightarrow} (x+y) dr + \int_{\nwarrow} (x+y) dr + \int_{\downarrow} (x+y) dr.$$ The tricky one is the slanting side, so let's do that last. The other two sides are $$\int_{\rightarrow} (x+y) dr = \int_0^1 (x + 0) dx = \left[ \frac{1}{2} x^2 \right]_0^1 = \frac{1}{2}$$ and $$\int_{\downarrow} (x+y) dr = \int_1^0 (0 +y) dy = \left[ -\frac{1}{2} y^2 \right]^1_0 = - \frac{1}{2}.$$ Observe that I didn't need to use any fancy machinery to compute these two - that's because these happen to lie along your coordinate axes. That's why in the first one my element is $dx$, i.e. the integral goes along the x axis (on which $y=0$, so I've set $y=0$ in the integrand), and similarly in the second one my element is $dy$. However here I integrate downwards, and to convey that I have to integrate from 1 to 0, as opposed to from 0 to 1 (recall that we are going anticlockwise).

To deal with the hypotenuse, parametrize it by $$\textbf{r}(t) = \left(x(t), y(t)\right) = \left(1-t, t\right).$$ Then I want to use this parameter $t$ to convert my integral along a line in two dimensions into a familiar integral in one dimension. Recall that by definition, $$\int_{\gamma} f dr = \int_a^b f(\textbf{r}(t))|\textbf{r}'(t)| dt. $$ In our case $a=0$ and $b=1$, and $$\left|\textbf{r}'(t)\right| = \left|\left(-1, 1\right)\right| = \sqrt{2},$$ so $$\int_{\nwarrow} (x+y) dr = \int_0^1 (1-t + t) \sqrt{2} dt = \sqrt{2}.$$ It remains to add the three parts, $$\int_{\gamma} (x+y) dr = \frac{1}{2} - \frac{1}{2} + \sqrt{2} = \sqrt{2}.$$

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  • $\begingroup$ Any chance of justifying the sign here - if we went clockwise the final answer should be $-\sqrt 2$ with the entire contribution coming from the diagonal. I'm having a hard time visualizing why upward to the left gives a positive result and downward to the right gives a negative result. $\endgroup$ – WW1 Nov 11 '15 at 21:44
  • $\begingroup$ It's convention. Along the diagonal $x+y$ always takes the value $1$, so the area under the curve $x+y$ is simply its value times the length of the curve, in this case $\sqrt{2}$. The direction in which we integrate doesn't change the area, of course, but it is useful to have a sense of direction for consistency. $\endgroup$ – onamoonlessnight Nov 11 '15 at 22:10
  • $\begingroup$ You can compare this to the familiar case of something like $$\int_0^1 x dx = \frac{1}{2}. $$ Obviously the area under the graph of $y(x) = x$ from $0$ to $1$ is $\frac{1}{2}$ in whichever direction we "add the infinitesimal rectangles", if that makes sense. But we need to have a sense of "doubling back on ourselves", so that if I integrate from $0$ to $1$ and then integrate backwards from $1$ to $\frac{1}{2}$, I should end up with an integral from $0$ to $\frac{1}{2}$. So my integration backwards should have a negative contribution; this is the essence behind $$\int_a^b = - \int_b^a.$$ $\endgroup$ – onamoonlessnight Nov 11 '15 at 22:15
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    $\begingroup$ @onemoonlessnight Perhaps the whole issue of clockwise versus counterclockwise is only relevant when dealing with line integrals of vector functions. Since $f$ is a scalar function we should be taking the absolute values of the integrals yielding an answer of $1+\sqrt 2$, which would accurately reproduce the area of the surface between the triangle and the plane. $\endgroup$ – WW1 Nov 12 '15 at 2:32

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