Question

find the area of the surface obtained by rotating the curve about the y-axis

Answer 1

y=x^2/4 - lnx/2 1<x<2

Answer 2

please help

Answer 3

hello

Answer 4

Is the curve \(y=\dfrac{x^2}{4}-\ln\dfrac{x}{2}\) ?

Answer 5

yes

Answer 6

could you tell me step by step on how to do this

Answer 7

the general form is
$$\LARGE \int 2\pi x~ ds$$

Answer 8

okay

Answer 9

Usually, plotting the curve helps with visualizing the dimensions of each cross-section, but that might be somewhat difficult given the form of this curve...

When you revolve a curve about a given axis, you're essentially drawing an infinite set of circles with radii defined by the distance between the curve and the axis of revolution. This means that, depending on the orientation of axis, you will have radii in terms of \(x\) (if the axis is a horizontal line) or \(y\) (if the axis is vertical).

The first thing to do then, since the axis is vertical, is to describe the curve as a function \(x(y)\). Plotting it in Mathematica reveals that the curve is not one-to-one in the interval \([1,2]\), so you may have to split up the domain and set up two integrals for the surface area.

Answer 10

so you have two options
$$\LARGE \int_{a}^{b} 2\pi x \sqrt{1 + [f ' (x)]^2}~dx \\

\LARGE \int_{c}^{d} 2\pi g(y) \sqrt{1 + [g ' (y)]^2}~dy

$$

Answer 11

the first one i think would be easier to do

Answer 12

@perl

Answer 13

Here's a plot of the region

Answer 14

Revolving about the y-axis means you need to be able to express the radius of each circle in terms of the variable \(y\), so you will have to split up the domain \(1\le x\le2\) to \((1\le x\le\sqrt2)\cup(\sqrt 2\le x\le2)\).

You also need to find the inverse function of \(y(x)\), which doesn't look like a simple task... Are you sure that the given function, axis, interval are correct?

Answer 15

https://www.desmos.com/calculator/p4y7gzfaxu

Answer 16

I think you have a typo in your question, this is a nasty integral because of the way you defined f(x)

Answer 17

it should be :
$$ \LARGE f(x) = \frac {x^2}{4} - \frac{ln(x)}{2} $$

Answer 18

We have
$$ \Large \int_{a}^{b} 2\pi x \sqrt{1 + (f ' (x))^2}~dx \\
\Large f(x)=\dfrac{x^2}{4}-\frac {ln(x)}{2} \\
\Large f ' (x) = \frac{x}{2}- \frac{1}{2x} \\
\Large \int_{1}^{2} 2\pi x \sqrt{1 + (\frac{x}{2}- \frac{1}{2x})^2}~dx $$

Answer 19

$$
\Large \int_{1}^{2} 2\pi x \sqrt{1 + (\frac{x}{2}- \frac{1}{2x})^2}~dx \\

\Large \int_{1}^{2} 2\pi x \sqrt{1 + \frac{1}{4}x^2-\frac{1}{2}+ \frac{1}{4x^2}}~dx \\

\Large \int_{1}^{2} 2\pi x \sqrt{\frac{1}{2} + \frac{1}{4}x^2+ \frac{1}{4x^2}}~dx \\

\Large \int_{1}^{2} 2\pi x \sqrt{\frac{x^4+2x^2+1}{4x^2}}~dx\\

\Large \int_{1}^{2} 2\pi x \sqrt{\frac{(x^2+1)^2}{4x^2}}~dx \\

\Large \int_{1}^{2} 2\pi x \frac{(x^2+1)}{2x}~dx \\
\Large \int_{1}^{2} \pi (x^2+1)~dx \\

\Large \pi \int_{1}^{2} (x^2+1)~dx \\

\Large \pi ~(\frac{x^3}{3} + x ) \LARGE|_{1}^{2} \\
\Large \pi((\frac{2^3}{3} +2) -(\frac{1^3}{3}+1))\\
\Large \pi (\frac{14}{3}-\frac{4}{3})\\
\Large \frac{10}{3} \pi

$$

Answer 20

@SithsAndGiggles

can you check Mathematica for the surface area of revolution for the revised function
$$ \large f(x) = \frac{x^2}{4} - \frac{ln(x)}{2} $$ on the interval (1,2)

Answer 21

about the y axis