Using an appropriate change of variables, calculate
Using an appropriate change of variables, calculate the area bounded by the curve: $$ \frac{x^2}4+\frac{y^2}9 = \frac{x^2+y^2}{25}. $$
Explanation of Method
A note on the equation as literally stated: multiplying through by $900$ (the LCM of $4,9,25$) gives $225x^2+100y^2=36x^2+36y^2$, i.e. $189x^2+64y^2=0$ — since both coefficients are positive, this is satisfied only at the origin, so the literal equation encloses no area at all. This is almost certainly a transcription slip for the natural "Cassini-oval-type" curve obtained by squaring the left side, $\left(\dfrac{x^2}4+\dfrac{y^2}9\right)^2=\dfrac{x^2+y^2}{25}$, which does bound a genuine closed region and is the version solved below (following the same reinterpretation used in the source).
- Substitute elliptical coordinates $x=2r\cos\theta,\,y=3r\sin\theta$.
- Transform the (reinterpreted) curve into a polar equation $r^2(\theta)$.
- Integrate over the full range $0\leqslant\theta\leqslant2\pi$.
Step 1 · Elliptical coordinates
Let $x=2r\cos\theta,\ y=3r\sin\theta$. Then $\dfrac{x^2}4+\dfrac{y^2}9=r^2$ and $x^2+y^2=r^2(4\cos^2\theta+9\sin^2\theta)$. The Jacobian is:
$$ \frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} 2\cos\theta & -2r\sin\theta \\ 3\sin\theta & 3r\cos\theta \end{vmatrix} = 6r\cos^2\theta+6r\sin^2\theta = 6r. $$
Step 2 · Transform the curve
With $\left(\dfrac{x^2}4+\dfrac{y^2}9\right)^2=\dfrac{x^2+y^2}{25}$:
$$ (r^2)^2 = \frac{r^2(4\cos^2\theta+9\sin^2\theta)}{25} \quad\Longrightarrow\quad r^2=\frac{4\cos^2\theta+9\sin^2\theta}{25}\quad(r\neq0). $$
This is real and positive for every $\theta$, so the curve is a single closed loop encircling the origin once.
Step 3 · Set up and evaluate the area integral
With area element $dx\,dy=6r\,dr\,d\theta$:
$$ A = \int_0^{2\pi}\int_0^{r(\theta)}6r'\,dr'\,d\theta = 3\int_0^{2\pi}r^2(\theta)\,d\theta = \frac3{25}\int_0^{2\pi}\big(4\cos^2\theta+9\sin^2\theta\big)\,d\theta. $$
Using $\displaystyle\int_0^{2\pi}\cos^2\theta\,d\theta=\int_0^{2\pi}\sin^2\theta\,d\theta=\pi$:
$$ A = \frac3{25}(4\pi+9\pi) = \frac3{25}\cdot13\pi = \frac{39\pi}{25}. $$
Additional Notes
- As noted above, the equation exactly as stated, $\frac{x^2}4+\frac{y^2}9=\frac{x^2+y^2}{25}$, reduces to $189x^2+64y^2=0$ and encloses no area (a degenerate point at the origin). The area computed here is for the natural reinterpretation $\left(\frac{x^2}4+\frac{y^2}9\right)^2=\frac{x^2+y^2}{25}$, matching the source's own resolution of this ambiguity.
- Verified numerically: direct quadrature of $3\int_0^{2\pi}\frac{4\cos^2\theta+9\sin^2\theta}{25}\,d\theta$ gives $\approx4.900885$, matching $\frac{39\pi}{25}\approx4.900885$.
Result
The area of the reinterpreted curve is: