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I'm working through this problem,

Compute a surface area by integration to show that if the plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ intersects the axes at points $A,B,C$ then Area of Triangle $= \sqrt{b^2c^2+c^2a^2+a^2b^2}$

and keep hitting a bump along the way I'm not sure how to overcome.. So far I have:

Let $z=f(x,y)=c\left(1-\frac{x}{a}-\frac{y}{b}\right)$ and thus used:

$$A(S)=\iint_{S}dS=\iint_{D}\sqrt{1+\left(\frac{\partial{z}}{\partial{x}}\right)^2+\left(\frac{\partial{z}}{\partial{y}}\right)^2}dA=\iint_{D}\sqrt{b^2c^2+c^2a^2+a^2b^2}dA$$

Now when considering D (the projection of $S$ onto the $xy$ plane, I have found a triangle with vertices $(0,0), (a,0), (0,b)$. Thus, I had limits of integration as: $$0\leq x\leq a$$$$0 \leq y \leq b\left(1-\frac{x}{a}\right)$$

My issue is that evaluating this I am obtaining $$\int_{0}^{a}\int_{0}^{b\left(1-\frac{x}{a}\right)}\sqrt{b^2c^2+c^2a^2+a^2b^2} dydx = \frac{ab}{2}\sqrt{b^2c^2+c^2a^2+a^2b^2}$$

I can see if $a=b=1$ then the desired result of $\frac{1}{2}\sqrt{b^2c^2+c^2a^2+a^2b^2}$ is obtained, but I'm not sure how to state my final result from this, or if I have made an error in the integral itself.

Any help massively appreciated!

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    $\begingroup$ $\sqrt{1+\left(\frac{\partial{z}}{\partial{x}}\right)^2+\left(\frac{\partial{z}}{\partial{y}}\right)^2}=\sqrt{1+(-\frac{c}{a})^2+(-\frac{c}{b})^2}=\frac{1}{ab}\sqrt{a^2b^2+c^2b^2+c^2a^2}$ - that is where the $ab$ in the denominator comes from, to eventually cancel $ab$ in the numerator. $\endgroup$ – Stinking Bishop Nov 6 '20 at 12:16
  • $\begingroup$ Ahh thank you!! Just my own algebra failing me then.. :) $\endgroup$ – 2307 Nov 6 '20 at 12:26
  • $\begingroup$ De Gua, see en.wikipedia.org/wiki/De_Gua%27s_theorem, is of help. So the claim seems to wrong by factor $2$ so your calculation seems right ... $\endgroup$ – Michael Hoppe Nov 6 '20 at 14:40
  • $\begingroup$ $$area=\frac{1}{2} \sqrt{a^2 b^2+a^2 c^2+b^2 c^2}$$ $\endgroup$ – Raffaele Nov 6 '20 at 17:53
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    $\begingroup$ Another solution: the vertices of the triangle are $(a,0,0),(0,b,0),(0,0,c)$ so two edge vectors of the triangle are $(-a,b,0),(-a,0,c)$. Now, if you take the cross product of those two you get $(bc, ac, ab)$ (and as a double check, this is indeed a scalar multiple of the normal vector $(1/a, 1/b, 1/c)$ to the plane). And finally, the area of the triangle is one half of the length of the cross product. $\endgroup$ – Daniel Schepler Nov 7 '20 at 0:00
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The three coordinate planes and the given plane $\pi$ determine a simplex $T$ in the first octant. The volume of $T$ can be computed in two ways: $${\rm vol}(T)={1\over6}abc={1\over3}A(S)h\ ,$$ where $h$ is the height of $T$ with respect to $S$. It follows that $${\rm area}(S)={abc\over 2h}\ .$$ In order to determine $h$ we intersect the plane normal $t\mapsto t\left({1\over a},{1\over b},{1\over c}\right)$ with $\pi$. When $P$ is the resulting point we have $$h=|OP|={ab c\over\sqrt{b^2c^2+c^2a^2+a^2b^2}}\ ,$$ so that $${\rm area}(S)={1\over2}\sqrt{b^2c^2+c^2a^2+a^2b^2}\ .$$

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