polar curves integration with $(dx)^2$?

Question

Find the area which lies between the x-axis and the curve $x = sin(t)$, $y = sin(t)cos(t)$, where $0 \le t \le \pi/2$

I was able to sketch a graph in the x-y coordinate plane by making a table of $t$, $x$, and $y$, and I know that the next step is to integrate, but I'm a little confused as to how to go about it.

Would it be $\int_{t_0}^{t_1} y(t) dx = \int_{0}^{\pi/2} {\sin(t) \cos(t)} dx$, where $dx = \cos(t) dt$? I think that should be right enough, but my friend says this could also work: $\int_{x_0}^{x_1} y dx$, where $y = x dx$ and so it'd be $\int_{0}^{1} x d^2x$—but is the latter a valid integral? Or is there another way?

Answer 1

Suppose the Cartesian form of the integral is $y=f(x),\,x\in[a,b]$. Let the parametrization $x=g(t)$ and $y=f(g(t)) = h(t)$ with $t\in[\alpha,\beta]$. It follows that $dx=g'(t)dt$ and

$$A=\int_{a=f(\alpha)}^{b=f(\beta)}f(x)\,dx=\int_{\alpha}^\beta h(t)g'(t)dt.$$

For your example, $g(t)=\sin(t)$ and $h(t)=\sin(t)\cos(t)$ with $t\in[0,\pi/2]$, so

$$A=\int_0^{\pi/2}\sin(t)\cos^2(t)\,dt\to\text{$u=\cos(t)$}\to\int_0^1u^2\,du=\frac13.$$

You could also subsitute $x(t)$ into $y(t)$, yielding $y=x\sqrt{1-x^2},\,$ $x\in[0,1]$ which when integrating yields the same answer of $1/3$.

Answer 2

You must be able to visualize what the expression means.

Basically, if you add up the area of the infinitesimal rectangles, it gives you the Area.

Area of one such rectangle is $dA=y\left(x\right)dx$. Hence the integral must always be $A=\int_{ }^{ }dA=\int_{ }^{ }y\left(x\right)dx$.

Rest is all mathematical manipulation. If you wish to solve this in terms of $t$, then convert both $x$ and $y$ in terms of $t$. If you wish to do it in terms of $x$, write $y$ in terms of $x$.

In the $t$ world,

$$dx=\cos\left(t\right)dt$$ $$y=\sin\left(t\right)\cos\left(t\right)$$ $$A=\int_{0}^{\frac{\pi}{2}}\sin\left(t\right)\cos^{2}\left(t\right)dt=\frac13$$

In the $x$ world,

$$y=\sin\left(t\right)\cos\left(t\right)=x\left(\sqrt{1-x^{2}}\right)$$ $$A=\int_{0}^{1}x\sqrt{1-x^{2}}dx=\frac13$$

Now, the thing about $(dx)^2$ is just some calculation mistake. In that way, it can be done as follows:

$$dx=\cos\left(t\right)dt$$ $$\cos\left(t\right)=\frac{dx}{dt}$$ $$y=\sin\left(t\right)\cos\left(t\right)=x\frac{dx}{dt}$$ $$A=\int_{ }^{ }x\cdot\frac{dx}{dt}\cdot dx$$

Again you need to change it either into $t$ world or into $x$ world. The results will be always the same. Just stick to basics. Cheers :)