Question

The given curve is rotated about the y-axis. Find the area of the resulting surface. y = 4 − x2, 0 ≤ x ≤ 3

Answer 1

Answer:

The area of the surface is 2.72 units

Step-by-step explanation:

The area of a curve y = (x) a ≤x ≤ b rotated about the y axis is given by:

$s=\int\limits^a_b {2\pi x\sqrt{1+(\frac{dy}{dx} )^2} } \, dx$

y = 4 - x²

dy / dx = -2x

(dy/dx)² = (-2x)² = 4x²

Hence:

$s=\int\limits^a_b {2\pi x\sqrt{1+(\frac{dy}{dx} )^2} } \, dx\\\\s=\int\limits^0_3 {2\pi x\sqrt{1+4x^2} } \, dx\\\\Let\ u = 1+4x^2\ \\\frac{du}{dx}=8x\\\frac{du}{8x}=dx\\\\Substituting\ value\ of\ u \ and\ dx:\\ \\s=\int\limits^0_3 {2\pi x\sqrt{u} } \, \frac{du}{8x} \\\\s=\frac{\pi }{4} \int\limits^0_3 {\sqrt{u} } \, du\\\\s=\frac{\pi }{4}|\frac{2}{3}u^{3/2}|_0^3\\\\s=\frac{\pi }{4}*\frac{2}{3}(5.196)\\\\s=2.72\ units$

The area of the surface is 2.72 units

Answer 2

The area of the resulting surface is $37.3437\pi$.

Curve:

A curve is an object that is similar to a line, but that does not have to be a straight line. A curve may be regarded as a trace left by a moving point.

Given:

$y=4-x^{2}$ , $0 \leq x\leq 3$

Surface area$=2\pi \int_{a}^{b}x\sqrt{1+\left ( \frac{dx}{dy} \right )^{2}dy}$

$x=\sqrt{4-y} \ , \ \frac{dx}{dy}=-\frac{1}{2\sqrt{4-y}}$

when $x=0 \ , \ y=4$

         $x=3 \ , \ y=-5$

Area of surface$=2\pi \int_{-5}^{4}\sqrt{4-y}\sqrt{1+\frac{1}{4\left ( y-4 \right )}dy}$

$=2\pi \int_{-5}^{4}\sqrt{4-y+\frac{4-y}{4\left ( y-4 \right )}dy} \\ =2\pi \int_{-5}^{4}\sqrt{4-y+\frac{1}{4}dy} \\ =\frac{2\pi }{2}\int_{-5}^{4}\sqrt{17-y} \ dy \\ =\pi \int_{-5}^{4}\sqrt{17-y} \ dy$

Solving with calculator:

$\Rightarrow \pi\left [ \frac{1}{6}\left ( 37\sqrt{37}-1 \right ) \right ]$

$=37.3437\pi$

Learn more about the topic Curve: https://brainly.com/question/8771120