Question

Find the surface area of that part of the plane 4𝑥+5𝑦+𝑧=8

4x+5y+z=8 that lies inside the elliptic cylinder (x²/100) + (y²/81) =1

Answer 1

$z=-4x-5y+8\\\frac{x^2}{100}+\frac{y^2}{81}\leqslant 1\\\sqrt{1+z{'_x}^2+z{'_y}^2}=\sqrt{1+4^2+5^2}=\sqrt{42}\\S=\iint_{\frac{x^2}{100}+\frac{y^2}{81}\leqslant 1}{\sqrt{42}dxdy}=\sqrt{42}S\left\{ \frac{x^2}{100}+\frac{y^2}{81}\leqslant 1 \right\} \\\frac{x^2}{100}+\frac{y^2}{81}\leqslant 1 is\,\,an\,\,ellipse\,\,with\,\,a=10,b=9, hence\\S\left\{ \frac{x^2}{100}+\frac{y^2}{81}\leqslant 1 \right\} =\pi ab=\pi \cdot 10\cdot 9=90\pi \\S=90\pi \sqrt{42}$