areadefinite integrals
I have the following curve:
$x^4=a^2(x^2-y^2)$
Prove that the area of its loop is $\frac{2a^2}{3}$.
My approach
This curve has four loops. So the required area should be:
$4\int_{0}^{a}\frac{x}{a}\sqrt{a^2-x^2} dx$
But, After solving, the area turned out to be $\frac{4a^2}{3}$. What am I doing wrong here?
Thanks
When they say
case 1:
When curve $y=f(X)$ lies above the $X$ axis
the area under curve is calculated using integration
- $\text{area} = \int y\,\text dx$ with some limits
They actually mean
case 1:
When curve $y=f(X)$ lies above the $X$ axis
the area between the curve and the $x$-axis is calculated using integration
- $\text{area} = \int y\,\text dx$ with some limits
So when you're integrating you're only getting half the area:
and naturally, you need to multiply that by $2$ to get all of it.
Also, remember that the actual integrand in question here is not $y^2 = 16x$, because that's not of the form $y = f(x)$ as requested from the cases. It's actually $y = 4\sqrt{x}$, which is only the upper half of the parabola I've sketched above. That makes it a bit more obvious why you don't get all of it; when integrating $y = 4\sqrt x$, your expressions have no way of knowing that it is only half of something bigger, and it especially can't know that the other half is the exact mirror image. So it does the best it can and finds the area down to the $x$-axis instead.
By symmetry, it is twice the area between the $y$-axis and the line $x=2$. The positive $x$-intercept is $x=1$, and on the interval $[0,1]$, the curve is below the $x$-axis, hence the total (geometric) area is $$2\biggl(\int_1^2y(x)\,\mathrm d x-\int_0^1y(x)\,\mathrm d x\biggl).$$
Best Answer
Your derivation seems correct, maybe the original question is simply asking for the area of one loop that is precisely
$$2\int_{0}^{a}\frac{x}{a}\sqrt{a^2-x^2} dx=\frac{2a^2}{3}$$