Caclulate the area of $ D=\{(x,y)|(x-y)^2+4x^2\leq R^2\}$

Question

Caclulate the area of $ D=\{(x,y)|(x-y)^2+4x^2\leq R^2\}$

Hint: Prove that for any continuous function $f$ $$\int_0^a\ dx \int_0^x f(y)\ dy = \int_0^a (a-x)f(x) \ dx$$

I’m honestly getting stuck lots of times at just sketching the regions even, does anyone have an idea of how to do this? And any tips on how to sketch a region in general and derive the limits of integration? Thanks.

Answer 1

Write $$2x=u,\quad y-x=v,\qquad {\rm resp.,}\qquad x={u\over2},\quad y={u\over2}+v\ .$$ Then $D$ is the linear image of $\hat D=\bigl\{(u,v)\bigm| u^2+v^2\leq R^2\bigr\}$. Since the Jacobian $x_uy_v-x_vy_u\equiv{1\over2}$ we have $${\rm area}(D)={1\over2}{\rm area}(\hat D)={\pi\over2}R^2\ .$$

Answer 2

Hint: The expression $$(x-y)^2+4x^2=5x^2-2xy+y^2=Ax^2+Bxy+Cy^2$$ where $B^2-4AC=-16<0$ is an ellipse, so $(x-y)^2+4x^2\leq R^2$ is the interior of it. One method is rotating the shape by $\alpha$ which $$\cot2\alpha=\dfrac{A-C}{B}$$

Edit: After rotation I got the new equation of ellipse as $$5(3-\sqrt{5})X^2+(3+\sqrt{5})Y^2=R^2$$ so $$S=\pi ab=\color{blue}{\dfrac{\pi}{2\sqrt{5}}R^2}$$

Answer 3

Write the equation of the curve as $$y^2-2xy+5x^2-R^2=0$$ this gives $$y=x\pm\sqrt{R^2-4x^2}.$$ So the required area is area between $y=x\pm \sqrt{R^2-4x^2}.$ from $x=-R/2$ to $x=R/2$.We get the required area $A$ by $$A=2\int_{=R/2}^{R/2} \sqrt{R^2-4x^2} dx==2\int_{0}^{R} \sqrt{R^2-z^2} dz=\pi R^2/2.$$ The given curve is an ellipse.