Find the area bounded by the following curve: (x²)/(a²) +
Find the area bounded by the following curve: $$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. $$
Explanation of Method
The substitution $x=au$, $y=bv$ maps the ellipse exactly onto the unit circle $u^2+v^2=1$, at the cost of a constant Jacobian factor $ab$. The area is then simply $ab$ times the area of the unit disk.
- Substitute $x=au$, $y=bv$ to map the ellipse to the unit circle.
- Compute the constant Jacobian of the transformation.
- Combine with the known area of the unit disk.
Step 1 · Elliptical substitution
Let $x=au$, $y=bv$. Then $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=u^2+v^2$, so the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}\leqslant1$ maps exactly onto the unit disk $u^2+v^2\leqslant1$.
Step 2 · Jacobian
$$ \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} a&0\\0&b \end{vmatrix} = ab, \qquad dx\,dy = ab\,du\,dv. $$
Step 3 · Compute the area
$$ A = \iint_{x^2/a^2+y^2/b^2\leqslant1} dx\,dy = \iint_{u^2+v^2\leqslant1} ab\,du\,dv = ab\cdot(\text{area of unit disk}) = ab\cdot\pi. $$
Additional Notes
- Equivalently, integrating directly in $(x,y)$: by symmetry, $A=4\displaystyle\int_0^a b\sqrt{1-x^2/a^2}\,dx$, and the substitution $x=a\sin\theta$ reduces this to $4ab\displaystyle\int_0^{\pi/2}\cos^2\theta\,d\theta=4ab\cdot\frac\pi4=\pi ab$, confirming the result by a second route.
- When $a=b=r$, the ellipse degenerates to a circle of radius $r$, and the formula correctly reduces to $A=\pi r^2$.
Result
The area is: