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

Wednesday, February 23, 2011

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 $\displaystyle\int\int_{{D}}\frac{{{2}{\left.{d}{x}\right.}{\left.{d}{z}\right.}}}{\sqrt{{{4}-{{x}}^{{2}}-{{z}}^{{2}}}}}$ 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 $\displaystyle{x}.$

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

This integral isn't so simple (inner integral is relatively simple), but at least we can compute it approximately. The answer is about 1.343.

Ninenotes

Wednesday, February 23, 2011

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

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