Bounded within what area?

Assume that the equation $y=f(x)$ is the boundary of the region in 1st quadrant. Then putting a tangent at any point must give me a line segment of length 2 units.Equation of tangent at $(x_0,f(x_0))$ is $y-f(x_0)=f'(x_0)(x-x_0)$ If $x=0$ , $y=f(x_0)-x_0f'(x_0)$ and if $y=0, x=x_0-\frac{f(x_0)}{f'(x_0)}$ . $(f(x_0)-x_0f(x_0))^2+(x_0-\frac{f(x_0)}{f'(x_0)})^2=2^2$ Now say $x_0=x$ And solve the differential equation. (Skipping a few steps) And the function is $f(x)=(2^{\frac{2}{3}}-x^{\frac{2}{3}})^{\frac{3}{2}}$ . This represents an astroid . Now all we have to do is integrate the function from 0 to 2 and then multiply it by 4 to get our answer.(Again skipping some steps). And the answer is 1.5

Best method to integrate is to use parametric coordinates followed by reduction formula. This has very less chance of making an error as you have a direct formula. Also , you could use trigonometric identities followed by integrating them but then there is a big risk of getting wrong answer.(I actually got the answer as 3 and then after many days found the error)

The side of a rhombuses is on the line $\frac{x}{2\cos\theta}+\frac{y}{2\sin\theta}=1$ .

To derive the envelops, we can tilt the line a little, $\mathrm{d}\theta$ , and the new line is $\frac{x}{2\cos (\theta+d\theta)}+\frac{y}{2\sin (\theta+d\theta)}=1$ .

Then we calculate out the joint point, $y=\frac{2\sin\theta \sin(\theta+\frac{\mathrm{d}\theta}{2})\sin (\theta+\mathrm{d}\theta)}{\cos \frac{d\theta}{2}}\approx 2 \sin^3\theta$ , $x\approx 2 \cos^3\theta$ .

For the area in the 1st quadrant, $S=\int_{0}^{2} y\mathrm{d}x=\int_{0}^{\frac{\pi}{2}}2\sin^3 \theta \mathrm{d}({2\cos^3 \theta})=\frac{3\pi}{8}$ . Then the total area is $\frac{3\pi}{2}$ .