# How do I calculate the area of the shape formed by this curve?

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?

• Do you mean the area under between $\pm\sqrt{x}(2-x)$ from $x=0$ to $x=2$?
– J.G.
Apr 25, 2020 at 21:09
• Yes, I believe that's what we are calculating (based on the graph). The question says "Calculate the area of shape fromed by curve $y^2 = x(x-2)^2$, so I couldn't deduce that without graphing it. Apr 25, 2020 at 21:12
• $$2\int_0^2 \sqrt{x}(2- x)dx= 2\int_0^2 x^{1/2}(2- x) dx= 2\int_0^2 (2x^{1/2}- x^{3/2})dx$$ Apr 25, 2020 at 21:15

$$y^2 = x(x-2)^2$$ 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 $$2 \int_0^2 \sqrt x \ |x-2| dx$$
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.