Area between $y^2=2ax$ and $x^2=2ay$ inside $x^2+y^2\le3a^2$

Question

I need to find the area between $y^2=2ax$ and $x^2=2ay$ inside the circle $x^2+y^2\le3a^2$. I know it's an integral but I can't seem to find the right one.

Answer 1

Disclaimer: This answer only works for positive values of $a$.

So we have the three equations, $y^2=2ax$, $x^2=2ay$, and $x^2+y^2\le3a^2$. If we rewrite these equations so that they all equal to $y$, we get $y=\sqrt{2ax}, y=x^2/2a,$ and $y\le\sqrt{3a^2-x^2}$, respectively. We can give names to these functions to make it easier: Let: $$P(x)=x^2/2a\tag{P for big parabola}$$ $$p(x)=\sqrt{2ax}\tag{p for little parabola}$$ $$C(x)=\sqrt{3a^2-x^2}\tag{C for circle}$$ With these equations we can now determine the limits of integration. First, we find the intersection of $P(x)$ and $p(x)$. $$x^2/2a=\sqrt{2ax}$$ $$x^4/4a^2=2ax$$ $$x^4=8a^3x$$ $$x(x^3-8a^3)=0$$ $$x=0\;\text{and}\;x=2a$$ So, we can now determine the area of the region enclosed by $P(x)$ and $p(x)$, by taking the integral from $0$ to $2$. Since $P(x)$ is less than $p(x)$, we get $$\int_0^{2a} \Big(p(x)-P(x)\Big)\,dx$$ which is equal to $$\int_0^{2a} \Big(\sqrt{2ax}\;- \;x^2/2a\Big)dx$$ $$=\frac13(2ax)^{3/2}-x^3/6a\;|\;0 \to 2a$$ $$=\frac43(2a^3-a^2)\tag1$$ This is the area enclosed by $p(x)$ and $P(x)$, but this area still includes the section outside $C(x)$, so we find the intersection of $P(x)$ and $C(x)$. $$x^2/2a=\sqrt{3a^2-x^2}$$ $$x^4/4a^2=3a^2-x^2$$ $$x^4=12a^4-4a^2x^2$$ $$x^4+4a^2x^2=12a^4$$ $$x^4+4a^2x^2+4a^4=16a^4$$ $$x^2+2a^2=\pm4a^2$$ $$x^2=-2a^2\pm4a^2$$ Since this value cannot be negative, we know that we can reject $-2a^2-4a^2$ and we get $$x=-2a^2+4a^2=2a^2$$ $$x=\sqrt{2}\,a$$ Now, we can find the area between $p(x)$ and $P(x)$ outside the boundaries of $C(x)$. $$\int_{\sqrt{2}a}^{2a} (p(x)-P(x))\;dx$$ $$=\frac13(2ax)^{3/2}-x^3/6a\;|\;\sqrt{2}\,a \to 2a$$ $$=\Bigg(\frac43(2a^3-a^2)\Bigg)-\Bigg(\frac13\Bigg(2^{9/4}a^3-2^{1/2}a^2\Bigg)\Bigg)\tag2$$ But we still have a section outside of the circle that we must calculate, between the intersection of $p(x)$ and $C(x)$, and $x=\sqrt{2}\,a$. So we find the intersection of $p(x)$ and $C(x)$. $$\sqrt{2ax}=\sqrt{3a^2-x^2}$$ $$2ax=3a^2-x^2$$ $$x^2+2ax=3a^2$$ $$x^2+2ax+a^2=4a^2$$ $$x+a=\pm2a$$ $$x=-a\pm2a$$ Since we know that the radius of the circle is only $\sqrt{3}\,a$, we also know that $x=-a-2a$ must be an extraneous solution. Thus, we have $$x=-a+2a=a$$ $$x=a$$ So, now we can find the area between $p(x)$ and $C(x)$ from $a$ to $\sqrt{2}\,a$. $$\int_{a}^{\sqrt{2}\,a} \Bigg(p(x)-C(x)\Bigg)\,dx$$ $$=\int_{a}^{\sqrt{2}\,a} \Bigg(\sqrt{2ax}-\sqrt{3a^2-x^2}\Bigg)\,dx$$ $$=\frac13(2ax)^{3/2}-\frac12\Bigg(x\sqrt{3a^2 - x^2}+3a^2\arctan{(x/\sqrt{3a^2-x^2})}\Bigg)\;|\;a \to \sqrt{2}\,a$$ $$=\Bigg(\Bigg[\frac13(2^{9/4}a^3)\Bigg]-\Bigg[\frac12\Bigg(2^{1/2}a^2+3a^2\arctan(\sqrt{2})\Bigg)\Bigg]\Bigg)-\Bigg(\Bigg[\frac13(2^{3/2}a^3)\Bigg]-\Bigg[\frac12\Bigg(2^{1/2}a^2+3a^2\arctan{(1/\sqrt{2})}\Bigg)\Bigg]\Bigg)$$ This is a little hairy, so we'll rearrange the values to get $$\frac{2^{9/4}a^3-2^{3/2}a^3}{3}+\frac{3a^2\arctan(1/\sqrt{2})-3a^2\arctan(\sqrt{2})}{2}\tag3$$ Now that we have calculated all the closed areas, we can combine them to get the closed area locked between $p(x)$,$P(x)$ and $C(x)$. $(1)$ gave us the area between $P(x)$ and $p(x)$, while $(2)$ and $(3)$, gave us the area between $P(x)$ and $p(x)$ outside of the circle $C(x)$. If we take $(1)-((2)+(3))$, we will get the area locked betweeen all three functions, the value of that expression is as follows: $$\Bigg(\frac43(2a^3-a^2)\Bigg)-\Bigg(\Bigg[\Bigg(\frac43(2a^3-a^2)\Bigg)-\Bigg(\frac13\Bigg(2^{9/4}a^3-2^{1/2}a^2\Bigg)\Bigg)\Bigg]+\Bigg[\frac{2^{9/4}a^3-2^{3/2}a^3}{3}+\frac{3a^2\arctan(1/\sqrt{2})-3a^2\arctan(\sqrt{2})}{2}\Bigg]\Bigg)$$ $$=\frac{3a^2\arctan(\sqrt{2})-3a^2\arctan(1/\sqrt{2})}{2}\tag4$$ Hopefully that answers your question.

Answer 2

You could probably use this integral:

$$ \int_0^a 2 \pi r \,dr $$

Answer 3

The parabolas will intersect the circle at the points $(a,\sqrt{2}a)$ and $(\sqrt{2}a,a)$ giving the following region:

So you want to find

$$ \int_0^a\sqrt{2ax}-\frac{x^2}{2a}\,dx+\int_a^{\sqrt{2}a}\sqrt{3a^2-x^2}-\frac{x^2}{2a}\,dx $$

In polar coordinates you can use the symmetry of the region and find the area by evaluating \begin{eqnarray} A&=&2\int_{\pi/4}^{\arctan(\sqrt{2})}\frac{r^2}{2}\,d\theta+2\int_{\arctan(\sqrt{2})}^{\pi/2}\frac{r^2}{2}\,d\theta\\ &=&\int_{\pi/4}^{\arctan(\sqrt{2})}3a^2\,d\theta+\int_{\arctan(\sqrt{2})}^{\pi/2}(2a\cot\theta\csc\theta)^2d\theta \end{eqnarray}

which can be finished by elementary means.

Answer 4

The entire figure scales proportionally to $a$, so we may set $a=1$ and multiply the area thus obtained by $a^2$ at the end. The desired area when $a=1$ is twice the area of the blue region $B$ above plus the area of the circular sector $C$ bounded by the black lines $y=\sqrt2x$ and $x=\sqrt2y$. $B$ is bounded by $y=\frac x{\sqrt2}$ from above and $y=\frac{x^2}2$ from below, so $$B=\int_0^{\sqrt2}\left(\frac1{\sqrt2}x-\frac12x^2\right)\,dx$$ $$=\left[\frac1{2\sqrt2}x^2-\frac16x^3\right]_0^{\sqrt2}=\frac{\sqrt2}6$$ The angle $\theta$ between the two black lines satisfies $$\tan\theta=\frac{\sqrt2-1/\sqrt2}{1+\sqrt2\cdot1/\sqrt2}=\frac{\sqrt2}4$$ which implies $\cos\theta=\frac{2\sqrt2}3$ and (since the sector radius is $\sqrt3$) $$C=\frac{r^2\theta}2=\frac32\cos^{-1}\frac{2\sqrt2}3$$ Finally, for a given $a$ the total area of the region in question is $$(2B+C)a^2=\left(\frac{\sqrt2}3+\frac32\cos^{-1}\frac{2\sqrt2}3\right)a^2$$ $$=0.981159\dots×a^2$$