To find the volume of the cap of a sphere with $x^2 + y^2 + z^2 \leq 26$, cut off by the plane $z = 5$ and restricted to the first octant, we can set up a triple integral in spherical coordinates. Here’s the breakdown:

Step-by-Step Setup

1. Convert to Spherical Coordinates:

   • In spherical coordinates:
     • $x = \rho \sin \phi \cos \theta$
     • $y = \rho \sin \phi \sin \theta$
     • $z = \rho \cos \phi$
   • The sphere $x^2 + y^2 + z^2 \leq 26$ translates to $\rho \leq \sqrt{26}$.
   • The plane $z = 5$ becomes $\rho \cos \phi = 5$, so $\phi = \arccos \left(\frac{5}{\sqrt{26}}\right)$.
   • The first octant limits $\theta$ from $0$ to $\frac{\pi}{2}$.

2. Determine the Limits:

   • $\rho$: Ranges from $0$ to $\sqrt{26}$.
   • $\phi$: Ranges from $0$ to $\arccos \left(\frac{5}{\sqrt{26}}\right)$.
   • $\theta$: Ranges from $0$ to $\frac{\pi}{2}$.

3. Volume Integral:

   • The volume element in spherical coordinates is $dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$.

Fill in the Blanks

• Coordinates: Spherical
• Limits:
  • $a = 0$, $b = \frac{\pi}{2}$ (limits for $\theta$)
  • $c = 0$, $d = \arccos \left(\frac{5}{\sqrt{26}}\right)$ (limits for $\phi$)
  • $e = 0$, $f = \sqrt{26}$ (limits for $\rho$)

Integral Expression for Volume

$\text{Volume} = \int_0^{\frac{\pi}{2}} \int_0^{\arccos \left(\frac{5}{\sqrt{26}}\right)} \int_0^{\sqrt{26}} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta$

Let me know if you need further details or have any questions.

Here are five related questions for further exploration:

1. How would the integral change if the plane was at $z = 4$ instead of $z = 5$?
2. How do the limits change if the cap is not restricted to the first octant?
3. What is the volume of the entire sphere using spherical coordinates?
4. How can we convert this volume calculation to cylindrical coordinates?
5. How would the setup differ for a sphere with a radius different from $\sqrt{26}$?

Tip: When setting up integrals in spherical coordinates, remember to adjust for the volume element $\rho^2 \sin \phi$.