# The part of the surface 2y+4z−x^2=5 that lies above the triangle with vertices (0, 0), (2, 0), and (2, 4) Find the area of the surface.

The part of the surface $$\displaystyle{2}{y}+{4}{z}−{x}^{{2}}={5}$$ that lies above the triangle with vertices $$\displaystyle{\left({0},{0}\right)},{\left({2},{0}\right)},{\quad\text{and}\quad}{\left({2},{4}\right)}$$ Find the area of the surface.

• Questions are typically answered in as fast as 30 minutes

### Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Usamah Prosser

$$\displaystyle{2}{y}+{4}{z}-{x}^{{2}}={5}$$
$$\displaystyle{f{{\left({x},{y}\right)}}}={\frac{{{5}}}{{{4}}}}+{\frac{{{x}^{{2}}}}{{{4}}}}-{\frac{{{2}{y}}}{{{4}}}}$$ $$\displaystyle{f}_{{x}}={\frac{{{x}}}{{{2}}}}.{f}_{{x}}=-{\frac{{{1}}}{{{2}}}}{0}\leq{x}\leq{2}$$

$$\displaystyle{S}=\int\int\sqrt{{{1}+{f}{x}^{{2}}+{f}{y}}}{d}{A}{0}\leq{y}\leq{2}{x}$$

$$dA=dydx$$

$$\displaystyle{S}={\int_{{{0}}}^{{{2}}}}{\int_{{{0}}}^{{{2}{x}}}}\sqrt{{{1}+{\frac{{{x}^{{2}}}}{{{4}}}}+{\frac{{{1}}}{{{4}}}}}}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$
$$\displaystyle{S}={\frac{{{1}}}{{{2}}}}{\int_{{{0}}}^{{{2}}}}{\int_{{{0}}}^{{{2}{x}}}}\sqrt{{{5}+{x}^{{2}}}}{\left.{d}{y}\right.}{\left.{d}{x}\right.}$$ $$\displaystyle{S}={\frac{{{1}}}{{{2}}}}{\int_{{0}}^{{2}}}{\left\lbrace{2}{x}\sqrt{{{5}+{x}^{{2}}}}\right\rbrace}{\left.{d}{x}\right.}$$
Let $$\displaystyle{5}+{x}^{{2}}={u}$$
$$\displaystyle{2}{x}{\left.{d}{x}\right.}={d}{u}$$ $$\displaystyle{S}={\frac{{{1}}}{{{2}}}}{\int_{{0}}^{{2}}}\sqrt{{{u}}}{d}{u}$$
$$\displaystyle{S}={\frac{{{1}}}{{{3}}}}{\left({5}+{X}^{{2}}\right)}^{{{\frac{{{3}}}{{{2}}}}}}{{\mid}_{{0}}^{{2}}}$$
$$\displaystyle{S}={\frac{{{1}}}{{{3}}}}{\left({27}-{5}\sqrt{{{5}}}\right)}$$