The task is to calculate the are of the shape formed by this curve:
$$y^2 = x(x-2)^2$$
Now, if I use Desmos to see what the graph looks like, I can do it. But if I get this question on my exam, I would not know what to do, because I could not visualise the curve. I'm in high school, so I'm not very familiar with such curves and I would like it if you can explain to me how to draw such a curve or perhaps, if it is possible to calculate the area without knowing what it looks like at all?
On
The given equation is equivalent to $y=\pm\sqrt{x}|2-x|$, so for real $y$ we need $x\ge0$. Since $y=0$ iff $x\in\{0,\,2\}$, and since as $x$ grows from $2$ to $\infty$ the two branches diverge as $y\to\pm\infty$, the desired shape is$$\int_0^2\sqrt{x}(2-x)dx-\left(-\int_0^2\sqrt{x}(2-x)dx\right)=2\int_0^2\sqrt{x}(2-x)dx.$$Now I've explained how to work out what area is desired without producing a diagram, I'll leave the calculus to you.
$$y^2 = x(x-2)^2$$ First, you may notice that this function is only defined for $x \geq 0$ assuming it is real-valued. Secondly, notice that it has roots $0$ with multiplicty $1$, and $2$ with multiplicity $2$. Also, it tends to $\infty$ as $x \to \infty$. We also have symmetry along the $x$-axis.
It may also be helpful to check where the derivative is $0$ for graphing functions like this (or any function, in fact). We have $$y’ = \frac{(x-2)(3x-2)}{2y} =0 \implies x=2 \, \text{or} \, x=\frac 23$$
Then it should be pretty clear that we have a maximum at $x=\frac 23$ and a minimum at $x=2$, and from here it is easy to deduce what area the question wants us to compute. In this case, it would be $$2 \int_0^2 \sqrt x \ |x-2| dx$$