Area enclosed between two curves

Question

Find the area enclosed by the curve in which the plane $z=2$ cuts the ellipsoid

$x^2/25 + y^2 + z^2/5 =1$

I tried to solve this by projecting the area on $xOy$ plane and I got the final answer as $\frac{\pi}{\sqrt{5}}$. Please correct me if I am wrong.

Answer 1

$$\frac{x^2}{25}+y^2+\frac{2^2}5=1$$

$$\frac{x^2}{25}+y^2=\frac15$$

$$\frac{x^2}{5}+\frac{y^2}{\frac15}=1$$

$$\left(\frac{x}{\sqrt5}\right)^2+\left(\frac{y}{\frac1{\sqrt5}}\right)^2=1$$

Hence the area is

$$\pi \sqrt5 \cdot \frac1{\sqrt5}=\pi$$

Remark:

We have $y=\pm \sqrt{\frac15-\frac{x^2}{25}}=\pm\frac1{\color{red}{\sqrt5}}\sqrt{1-\frac{x^2}5}$. If you have been working with $\frac15$ instead, you will gain an additional copy of $\frac1{\sqrt5}$.

Answer 2

We have that for $z=2$

$$\frac{x^2}{25} + y^2 + \frac{z^2}{5} =1 \implies \frac{x^2}{25} + y^2=\frac15\implies \frac{x^2}{({\sqrt 5})^2} + \frac{y^2}{(\frac{\sqrt 5}{5})^2}=1$$

which is an ellipse in the canonical form $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$, then recall that for an ellipse the area is $A=\pi ab$ to obtain $A=\pi$.