Blog: Determine the area "S" which the triangle "R" Project vertically upon the hemisphere. Check the attachment

Hello!

If a surface is given as an image of a scalar function $\displaystyle{y}={f{{\left({x},{z}\right)}}},$ defined on some region $\displaystyle{D}$ on $\displaystyle{\left({x},{z}\right)}$ plane, then the corresponding surface area is

$\displaystyle\int\int_{{D}}\sqrt{{{1}+{{\left(\frac{{\partial{f}}}{{\partial{x}}}\right)}}^{{2}}+{{\left(\frac{{\partial{f}}}{{\partial{z}}}\right)}}^{{2}}}}{\left.{d}{x}\right.}{\left.{d}{z}\right.}.$

We have $\displaystyle{f{{\left({x},{z}\right)}}}=\sqrt{{{4}-{{x}}^{{2}}-{{z}}^{{2}}}},$ so $\displaystyle\frac{{\partial{f}}}{{\partial{x}}}=-\frac{{x}}{{\sqrt{{{4}-{{x}}^{{2}}-{{z}}^{{2}}}}}}$ and $\displaystyle\frac{{\partial{f}}}{{\partial{z}}}=-\frac{{z}}{{\sqrt{{{4}-{{x}}^{{2}}-{{z}}^{{2}}}}}}.$ The expression under integral therefore is

$\displaystyle\sqrt{{{1}+\frac{{{x}}^{{2}}}{{{4}-{{x}}^{{2}}-{{z}}^{{2}}}}+\frac{{{z}}^{{2}}}{{{4}-{{x}}^{{2}}-{{z}}^{{2}}}}}}=\frac{{2}}{\sqrt{{{4}-{{x}}^{{2}}-{{z}}^{{2}}}}}.$

The problem is to express the double integral $\displaystyle\int\int_{{D}}\frac{{{2}{\left.{d}{x}\right.}{\left.{d}{z}\right.}}}{\sqrt{{{4}-{{x}}^{{2}}-{{z}}^{{2}}}}}$ ...

Hello!

$\displaystyle\sqrt{{{1}+\frac{{{x}}^{{2}}}{{{4}-{{x}}^{{2}}-{{z}}^{{2}}}}+\frac{{{z}}^{{2}}}{{{4}-{{x}}^{{2}}-{{z}}^{{2}}}}}}=\frac{{2}}{\sqrt{{{4}-{{x}}^{{2}}-{{z}}^{{2}}}}}.$

The problem is to express the double integral $\displaystyle\int\int_{{D}}\frac{{{2}{\left.{d}{x}\right.}{\left.{d}{z}\right.}}}{\sqrt{{{4}-{{x}}^{{2}}-{{z}}^{{2}}}}}$ as a sequential one-dimensional integral. Because of the symmetry we can integrate only by a half of the triangle and then multiply by $\displaystyle{2}.$ The integration region is from $\displaystyle-\frac{{1}}{{2}}$ to $\displaystyle{1}$ by $\displaystyle{z}$ and from $\displaystyle{0}$ to $\displaystyle\frac{{{1}-{z}}}{\sqrt{{{3}}}}$ by $\displaystyle{x}.$

$\displaystyle{A}={2}{\int_{{-\frac{{1}}{{2}}}}^{{1}}}{\left({\int_{{0}}^{{\frac{{{1}-{z}}}{\sqrt{{{3}}}}}}}\frac{{2}}{\sqrt{{{4}-{{x}}^{{2}}-{{z}}^{{2}}}}}{\left.{d}{x}\right.}\right)}{\left.{d}{z}\right.}.$