MHB Calculating the Area of Region Bound by a Curve and its Asymptotes

[Dethrone]

Find the area of the region between the given curve and its asymptotes.

$$y^2 = \frac{x^4}{4-x^2}
$$
The answer is $$4\pi$$

What I did was I integrated $$y = \frac{x^2}{\sqrt{4-x^2}}$$ but my answer is only $$2\pi$$. I'm thinking that by squaring the function, I lost one half the answer (the negative portion). Is that true, or is there another way to attempt this problem?

 

[MarkFL]

I would use the symmetry across both axes to give the area $A$ as:

$$A=4\int_0^2\frac{x^2}{\sqrt{4-x^2}}\,dx$$

 

[I like Serena]

Find the area of the region between the given curve and its asymptotes.

$$y^2 = \frac{x^4}{4-x^2}
$$
The answer is $$4\pi$$

What I did was I integrated $$y = \frac{x^2}{\sqrt{4-x^2}}$$ but my answer is only $$2\pi$$. I'm thinking that by squaring the function, I lost one half the answer (the negative portion). Is that true, or is there another way to attempt this problem?

Hi Rido!

Your method is fine.
It's just that the curve has a portion above the x-axis, which is the part that you've calculated, and also an identical part below the x-axis.

If you include the part below the x-axis, you'll have $4\pi$.

 

[Dethrone]

I understand what you're trying to say, and I came across that conclusion myself, but I can't arrive at it through my calculations.

Essentially, we need to compute $$\int_{-2}^{2} \, \frac{x^2}{\sqrt{4-x^2}}$$ + $$\int_{-2}^{2} \, \frac{-x^2}{\sqrt{4-x^2}}$$, or the area that is enclosed between the asymptotes of $$y=\frac{+/-x^2}{\sqrt{4-x^2}}$$

But if we start from:

$$y^2 = \frac{x^4}{4-x^2}$$, taking the square root from both sides gives us:
$$\mid y \mid = \frac{x^2}{\sqrt{4-x^2}}$$

If the function is absolute valued, how do I get the negative portion of the graph?

 

[I like Serena]

I understand what you're trying to say, and I came across that conclusion myself, but I can't arrive at it through my calculations.

Essentially, we need to compute $$\int_{-2}^{2} \, \frac{x^2}{\sqrt{4-x^2}}$$ + $$\int_{-2}^{2} \, \frac{-x^2}{\sqrt{4-x^2}}$$, or the area that is enclosed between the asymptotes of $$y=\frac{+/-x^2}{\sqrt{4-x^2}}$$

But if we start from:

$$y^2 = \frac{x^4}{4-x^2}$$, taking the square root from both sides gives us:
$$\mid y \mid = \frac{x^2}{\sqrt{4-x^2}}$$

If the function is absolute valued, how do I get the negative portion of the graph?

Erm... from
$$| y | = \frac{x^2}{\sqrt{4-x^2}}$$
it follows that
$$y = \pm \frac{x^2}{\sqrt{4-x^2}}$$

 

[Dethrone]

Absolute values confuse me. I used to think that to be true, until I came across this:

$$\mid \sin\left({x}\right) \mid = 1 -cos^2x
$$
so it follows that:
$$
\sin\left({x}\right)= \pm (1-cos^2x)$$, surely, $$\sin\left({x}\right)$$ can't be both at the same time?

 

[I like Serena]

Absolute values confuse me. I used to think that to be true, until I came across this:

$$\mid \sin\left({x}\right) \mid = 1 -cos^2x
$$
so it follows that:
$$
\sin\left({x}\right)= \pm (1-cos^2x)$$, surely, $$\sin\left({x}\right)$$ can't be both at the same time?

That should be:
$$\sin\left({x}\right)= \pm \sqrt{1-cos^2x}$$

And you are right that it cannot be both at the same time, but it can still be one or the other.
To clarify, what do you get if you substitute $x=\frac \pi 2$ respectively $x=-\frac \pi 2$?

In general, when considering if a solution is valid, we always need to go back to the original equation and verify whether the solution satisfies it.
At this point you can get interesting surprises. ;)

 

[Dethrone]

Well, putting either $$x=\frac \pi 2$$ or $$-\frac \pi 2$$ will return the same answer because cosine is an even function, but isn't it not very accurate to say
$$\sin\left({x}\right)= \pm \sqrt{1-cos^2x}$$ since it is only correct in a few regions?

Or does it not matter if your only purpose is to plug x-values in, and you're not trying to graph it...?

 

[I like Serena]

Well, putting either $$x=\frac \pi 2$$ or $$-\frac \pi 2$$ will return the same answer because cosine is an even function,

Cosine may be an even function, but sine isn't.

but isn't it not very accurate to say
$$\sin\left({x}\right)= \pm \sqrt{1-cos^2x}$$ since it is only correct in a few regions?

Or does it not matter if your only purpose is to plug x-values in, and you're no trying to graph it...?

Either the one or the other is valid, depending where you are in the domain.
It means you have to distinguish 2 cases.
And when you get to a final answer, you need to verify it against the original equation.

 

[Dethrone]

Arggg, absolute value really makes my head hurt.
$$
| y | = \frac{x^2}{\sqrt{4-x^2}}$$

This function is never negative, I just don't understand how it can end up as
$$y = \pm \frac{x^2}{\sqrt{4-x^2}}$$ where it can return negative and positive answers. By graphing, they're different.
I can see how they're true though, as $$ y^2 = 4$$ can have solutions $$\pm 2. $$

I've also seen scenarios in books where they rewrite
$$| y | = \frac{x^2}{\sqrt{4-x^2}}$$ as
$$y = \frac{x^2}{\sqrt{4-x^2}}$$ because no matter what values of x you put in, you end up with a positive value, so the absolute value can be omitted. i.e $$\ln\left({\mid x^2 + 1 \mid}\right)
$$ = $$\ln\left({x^2+1}\right)$$

Right now, I feel that $$
| y | = \frac{x^2}{\sqrt{4-x^2}}$$ can only equal $$y = \pm \frac{x^2}{\sqrt{4-x^2}}$$ if the purposes are for returning x and y solutions, but not for graphing. And once we get those x and y values, we verify them with the original equation but plugging them back in.

 

[I like Serena]

Arggg, absolute value really makes my head hurt.
$$
| y | = \frac{x^2}{\sqrt{4-x^2}}$$

This function is never negative,

The right hand side would indeed never be negative.
However, $y$ can still be negative.

I've also seen scenarios in books where they rewrite
$$| y | = \frac{x^2}{\sqrt{4-x^2}}$$ as
$$y = \frac{x^2}{\sqrt{4-x^2}}$$ because no matter what values of x you put in, you end up with a positive value, so the absolute value can be omitted.

Careful!

The absolute value can not be omitted.
Only if it is given (or is implicitly clear) that $y \ge 0$ can the absolute value be omitted.

i.e $$\ln\left({\mid x^2 + 1 \mid}\right)
$$ = $$\ln\left({x^2+1}\right)$$

Yes, in this case the absolute value can be omitted, because the argument of the absolute value function is non-negative.

 

[Dethrone]

Ah, thank you! It's all clear now. It's weird, because that's how I used to think about absolute value until I over-thought it.