Best solutions to - "The part of the surface 2y+4z−x2=5 that lies..."

College
Calculus and Analysis
Integral Calculus
Parametric equations, polar coordinates, and vector-valued functions

ddaeeric

Answered question

2021-09-14

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

Answer & Explanation

Usamah Prosser

Skilled2021-09-15Added 86 answers

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

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

$d A = d y d x$

$S = \int_{0}^{2} \int_{0}^{2 x} \sqrt{1 + \frac{x^{2}}{4} + \frac{1}{4}} d y d x$
$S = \frac{1}{2} \int_{0}^{2} \int_{0}^{2 x} \sqrt{5 + x^{2}} d y d x$ $S = \frac{1}{2} \int_{0}^{2} {2 x \sqrt{5 + x^{2}}} d x$
Let $5 + x^{2} = u$
$2 x d x = d u$ $S = \frac{1}{2} \int_{0}^{2} \sqrt{u} d u$
$S = \frac{1}{3} {(5 + X^{2})}^{\frac{3}{2}} ∣_{0}^{2}$
$S = \frac{1}{3} (27 - 5 \sqrt{5})$

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