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!
Best Answer
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 $${\rm area}(S)={1\over2}\sqrt{b^2c^2+c^2a^2+a^2b^2}\ .$$