Area bounded by a curve's loop

[Raghav Gupta]

Homework Statement

The area bounded by the loop of the curve $4y^2 = x^2(4-x^2)$ is in sq. units
7/3
8/3
11/3
16/3

Homework Equations

NA

The Attempt at a Solution

By putting x = 0 and x = 2 I am getting y = 0.
Getting complex y values after x exceeds 2.
I am not getting where the loop would form
I am not allowed to use graphing calculator to see shape.
Is there any better way than plotting for concavity of function, to determine loop, or something else?

 

[HallsofIvy]

The graph is a "lemniscate": http://en.wikipedia.org/wiki/Lemniscate- it looks like a two lobed airplane propeller. I would put this into polar coordinates.

 

[STEMucator]

Start by finding a polar representation of the given curve. You wind up with something like this:

$$r = f(\theta) = \sqrt{4\text{sec}^2(\theta) (1 - \text{tan}^2(\theta))}$$

Now you have a theorem which tells you the area of a polar region $R$ is given by:

$$\frac{1}{2} \int_a^b f^2(\theta) \space d \theta$$

 

[Raghav Gupta]

The graph is a "lemniscate": http://en.wikipedia.org/wiki/Lemniscate- it looks like a two lobed airplane propeller. I would put this into polar coordinates.

Start by finding a polar representation of the given curve. You wind up with something like this:

$$r = f(\theta) = \sqrt{4\text{sec}^2(\theta) (1 - \text{tan}^2(\theta))}$$

Now you have a theorem which tells you the area of a polar region $R$ is given by:

$$\frac{1}{2} \int_a^b f^2(\theta) \space d \theta$$

But you may have used graphing calculator to see the shape as a lemniscate.
How to know shape if you have not used the calculator?

 

[Raghav Gupta]

Start by finding a polar representation of the given curve. You wind up with something like this:

$$r = f(\theta) = \sqrt{4\text{sec}^2(\theta) (1 - \text{tan}^2(\theta))}$$

Now you have a theorem which tells you the area of a polar region $R$ is given by:

$$\frac{1}{2} \int_a^b f^2(\theta) \space d \theta$$

I think for finding area once we know graph can easily do,
$$ \int_0^2 \sqrt{x^2(4-x^2)}dx $$

 

[STEMucator]

There are several ways to graph $r = f(\theta)$. The one I usually use is to plot $f(\theta)$ in the Cartesian plane for several radial arguments. This will allow you to trace the graph in the $r - \theta$ plane and obtain the limits of integration. These are the limits for the theorem in post #3.

The theorem actually makes it quite easy to integrate in polar co-ordinates as well (which is what the exercise intended I believe). The substitution $u = \text{tan}^2(\theta)$ makes the rest easy.

 

[Ray Vickson]

There are several ways to graph $r = f(\theta)$. The one I usually use is to plot $f(\theta)$ in the Cartesian plane for several radial arguments. This will allow you to trace the graph in the $r - \theta$ plane and obtain the limits of integration. These are the limits for the theorem in post #3.

The theorem actually makes it quite easy to integrate in polar co-ordinates as well (which is what the exercise intended I believe). The substitution $u = \text{tan}^2(\theta)$ makes the rest easy.

Yes, but the OP's alternative $\int \sqrt{x^2(4-x^2)} \, dx = \int x \sqrt{4 - x^2} \, dx$ is even easier.

 

[Raghav Gupta]

There are several ways to graph $r = f(\theta)$. The one I usually use is to plot $f(\theta)$ in the Cartesian plane for several radial arguments. This will allow you to trace the graph in the $r - \theta$ plane and obtain the limits of integration. These are the limits for the theorem in post #3.

The theorem actually makes it quite easy to integrate in polar co-ordinates as well (which is what the exercise intended I believe). The substitution $u = \text{tan}^2(\theta)$ makes the rest easy.

I have not read that polar coordinate things till now.

Yes, but the OP's alternative $\int \sqrt{x^2(4-x^2)} \, dx = \int x \sqrt{4 - x^2} \, dx$ is even easier.

Can you tell me the answer for below quotation?

But you may have used graphing calculator to see the shape as a lemniscate.
How to know shape if you have not used the calculator?

 

[Ray Vickson]

I have not read that polar coordinate things till now.

Can you tell me the answer for below quotation?

See, eg., http://en.wikipedia.org/wiki/Lemniscate . Click on the link to "Lemniscate of Gerono".

 

[Raghav Gupta]

See, eg., http://en.wikipedia.org/wiki/Lemniscate . Click on the link to "Lemniscate of Gerono".

But suppose we do not know these things.
Doing some basics. We have to find only area of any curve . For example here I found y value 0 when x is 0 and y value 0 when x is 2. I am getting complex number value when taking x value more than 2.
From this how I can arrive at shape?

 

[Ray Vickson]

But suppose we do not know these things.
Doing some basics. We have to find only area of any curve . For example here I found y value 0 when x is 0 and y value 0 when x is 2. I am getting complex number value when taking x value more than 2.
From this how I can arrive at shape?

Plot the curve.

Do not tell me that you are not allowed to plot things; of course you are---you just cannot turn in the plot as a final solution for mark! Nobody can stop you from making plots in the privacy of your own room. The plot does not have to be very accurate; it is enough that it supplies a rough idea of the shape.

Alternatively: to see what $y = c x \sqrt{4 - x^2}$ looks like look at $y^2 = c^2 x^2(4-x^2)$. As a function of $t = x^2$, the rhs is $c^2 t(4-t)$, which rises from 0 at $t = 0$ to a maximum at $t = 2$ and then falls to 0 again at $t = 4$. So $y^2$ rises fro 0 to a maximum at $x = \sqrt{2}$, then falls to 0 again at $x = 2$. Wherever $y^2$ is rising, so is $y$, and wherever $y^2$ is falling, so is $y$. Therefore, $y$ rises from 0 to a maximum at $x = \sqrt{2}$ then falls to zero again.

The lower branch $y = - c x \sqrt{4 - x^2}$ falls from 0 to a minimum then rises again to 0.

 

[STEMucator]

Really though, polar co-ordinates might help you plot this easier. You shouldn't be lazy about doing plots, they are tedious, but a necessity.

A picture is worth a million words.

Speaking of pictures, maybe this one will help you understand what I meant in post #6:

The function $r = f(\theta)$ might look simpler, but the procedure is the same.

 

[Raghav Gupta]

Thanks Zondrina, Ray, Halls although Zondrina I don't know anything about polar coordinate system.
In school level I suppose it is not taught.

 

[certainly]

I don't know anything about polar coordinate system.
In school level I suppose it is not taught.

It is usually taught in high school. Maybe your teacher thought it was too obvious to mention. See here.
[EDIT:- you just need to know what $r$ and $\theta$ (and $\phi$ if your working in 3 dimensions) are, that's all there is to it really.]

 

[Raghav Gupta]

It is usually taught in high school. Maybe your teacher thought it was too obvious to mention. See here.
[EDIT:- you just need to know what $r$ and $\theta$ (and $\phi$ if your working in 3 dimensions) are, that's all there is to it really.]

What is the use of polar coordinates when we can do the thing of post 5 easily?

 

[certainly]

What is the use of polar coordinates when we can do the thing of post 5 easily?

Nothing. But polar coordinates become extremely useful when you try to describe curves like cycloids and trochoids. A lot of their properties become a lot easier to prove, describe and visualize in polar coordinates. I'm not sure what level you're at, but the 4th chapter of Courant's analysis book has an excellent section on the theory of plane curves. If you read it, the use of polar coordinates will become immediately clear.