Find the surface area bounded by (x cos alpha+y sin

Find the surface area bounded by $(x\cos\alpha+y\sin\alpha)^2+z^2=a^2$, $x>0,\,y>0,\,z>0$.

Step 1 · Lateral cylindrical surface

Parametrize $u=a\cos t,\,w=a\sin t$ ($t\in(0,\pi/2)$ so $z=w>0$), $v=s$. The constraints $x>0,y>0$ bound $s\in(-a\cos t\tan\alpha,\,a\cos t\cot\alpha)$, an interval of length $\Delta s=\dfrac{2a\cos t}{\sin2\alpha}$. The surface-element magnitude $\|\mathbf r_t\times\mathbf r_s\|=a$ (a standard cylinder fact), so:

$$ S_{\text{cyl}} = a\int_0^{\pi/2}\Delta s\,dt = \frac{2a^2}{\sin2\alpha}\int_0^{\pi/2}\cos t\,dt = \frac{2a^2}{\sin2\alpha}. $$

Step 2 · The two elliptical faces

On $x=0$: the cylinder becomes $y^2\sin^2\alpha+z^2\le a^2$ (a quarter-ellipse, $y,z\ge0$, semi-axes $a/\sin\alpha$ and $a$):

$$ A_{x=0} = \frac\pi4\cdot\frac a{\sin\alpha}\cdot a = \frac{\pi a^2}{4\sin\alpha}. $$

On $y=0$, symmetrically:

$$ A_{y=0} = \frac{\pi a^2}{4\cos\alpha}. $$

Step 3 · The overlooked $z=0$ face

At $z=0$, the solid's cross-section is $\{x>0,y>0,\,u\le a\}$ where $u=x\cos\alpha+y\sin\alpha\ge0$ automatically — a genuine triangle bounded by the axes and the line $x\cos\alpha+y\sin\alpha=a$, with $x$-intercept $a/\cos\alpha$ and $y$-intercept $a/\sin\alpha$:

$$ A_{z=0} = \frac12\cdot\frac a{\cos\alpha}\cdot\frac a{\sin\alpha} = \frac{a^2}{\sin2\alpha}. $$

Summing all four pieces:

$$ S = S_{\text{cyl}}+A_{x=0}+A_{y=0}+A_{z=0} = \frac{2a^2}{\sin2\alpha}+\frac{a^2}{\sin2\alpha}+\frac{\pi a^2}{4\sin\alpha}+\frac{\pi a^2}{4\cos\alpha} = \frac{3a^2}{\sin2\alpha}+\frac{\pi a^2(\sin\alpha+\cos\alpha)}{2\sin2\alpha}. $$

Combining over the common denominator $\sin2\alpha$:

$$ S = \frac{a^2}{\sin2\alpha}\left[3+\frac\pi2(\sin\alpha+\cos\alpha)\right]. $$

Additional Notes

• A note on the source computation: the source correctly identified that the closed solid's boundary needs a face on each of $x=0$, $y=0$, and $z=0$ (in addition to the lateral cylinder), and correctly computed the two elliptical faces. However, for the $z=0$ face it incorrectly concluded the area was zero, reasoning that "on $z=0$, the cylinder gives $u=\pm a$... this is a line (zero area)." This conflates the boundary curve of the lateral cylindrical surface at $z=0$ (which is indeed just the single curve $u=a$, since that surface only has $t=0$ mapping to $z=0$) with the actual flat face needed to close the solid at $z=0$, which is the full 2D interior cross-section $u\le a$ (not just the boundary $u=a$) intersected with $x,y>0$ — a genuine triangle of area $a^2/\sin2\alpha$, as derived above. Omitting this face understated the total surface area by exactly that amount.
• Verified numerically (e.g. $\alpha=\pi/6,\,a=1.7$): $S_{\text{cyl}}\approx6.6742$, $A_{x=0}\approx4.5396$, $A_{y=0}\approx2.6209$, $A_{z=0}\approx3.3371$ (the face the source omitted), total $\approx17.1718$, matching the corrected closed form exactly — versus $\approx13.8347$ if that face is left out, matching the source's (incomplete) claimed total.