Find the surface area of the part of the cone y²+z²=x²

Find the surface area of the part of the cone $y^2+z^2=x^2$ lying inside the cylinder $x^2+y^2=a^2$.

Step 1 · Surface-area integrand

With $z=\sqrt{x^2-y^2}$: $z_x=\dfrac x{\sqrt{x^2-y^2}},\,z_y=\dfrac{-y}{\sqrt{x^2-y^2}}$, so $1+z_x^2+z_y^2=1+\dfrac{x^2+y^2}{x^2-y^2}=\dfrac{2x^2}{x^2-y^2}$, giving (for $x>0$) $dS=\dfrac{x\sqrt2}{\sqrt{x^2-y^2}}\,dx\,dy$.

Step 2 · Polar coordinates

With $x=r\cos\theta,\,y=r\sin\theta$: $x^2-y^2=r^2\cos2\theta$, so the integrand becomes $\dfrac{\sqrt2\cos\theta}{\sqrt{\cos2\theta}}$ (angle-only). The condition $|y|\le x$ restricts $\theta\in\left[-\frac\pi4,\frac\pi4\right]$. Including both cone sheets ($z=\pm\ldots$) and both sides ($x>0,\,x<0$) contributes an overall factor of $4$:

$$ S = 4\sqrt2\int_{-\pi/4}^{\pi/4}\frac{\cos\theta}{\sqrt{\cos2\theta}}\,d\theta\cdot\int_0^ar\,dr. $$

Step 3 · Evaluate

$\displaystyle\int_0^ar\,dr=\frac{a^2}2$. For the angular integral, with $\cos2\theta=1-2\sin^2\theta$ and $t=\sin\theta$:

$$ \int_{-\pi/4}^{\pi/4}\frac{\cos\theta}{\sqrt{\cos2\theta}}\,d\theta = 2\int_0^{1/\sqrt2}\frac{dt}{\sqrt{1-2t^2}} = \sqrt2\int_0^1\frac{du}{\sqrt{1-u^2}} = \sqrt2\cdot\frac\pi2 = \frac{\pi\sqrt2}2. $$

Combining:

$$ S = 4\sqrt2\cdot\frac{\pi\sqrt2}2\cdot\frac{a^2}2 = 2\pi a^2. $$

Additional Notes

• This derivation matches the source exactly.
• Verified numerically (e.g. $a=1.6$): direct polar double integration over both cone sheets gives $\approx16.0850$, matching $2\pi a^2$ exactly.