To solve the problem using triple integrals and determine the volume bounded by the given cone and sphere, we need to establish the correct limits for integration.

Step 1: Understand the geometry

1. Cone equation: $x^2 + y^2 = z^2$
   This is a cone symmetric about the $z$-axis, with its vertex at the origin.
2. Sphere equation: $x^2 + y^2 + z^2 = a^2$
   This is a sphere centered at the origin with radius $a$.

Step 2: Intersection of the cone and sphere

To find the intersection curve, substitute $x^2 + y^2 = z^2$ (from the cone) into $x^2 + y^2 + z^2 = a^2$ (from the sphere): $z^2 + z^2 = a^2 \quad \Rightarrow \quad 2z^2 = a^2 \quad \Rightarrow \quad z^2 = \frac{a^2}{2} \quad \Rightarrow \quad z = \pm\frac{a}{\sqrt{2}}$ Thus, the cone intersects the sphere at $z = \frac{a}{\sqrt{2}}$ and $z = -\frac{a}{\sqrt{2}}$.

Step 3: Convert to spherical coordinates

To simplify the volume computation, use spherical coordinates: $x = \rho \sin\phi \cos\theta, \quad y = \rho \sin\phi \sin\theta, \quad z = \rho \cos\phi,$ where:

• $\rho$ is the radial distance,
• $\phi$ is the angle from the positive $z$-axis ($0 \leq \phi \leq \pi$),
• $\theta$ is the azimuthal angle ($0 \leq \theta < 2\pi$).

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

Step 4: Integration limits

• $\rho$: From the cone ($\rho = \frac{z}{\cos\phi} = \frac{\rho\cos\phi}{\cos\phi}$) to the sphere ($\rho = a$).
• $\phi$: From $0$ to $\frac{\pi}{4}$ (since the cone forms an angle of $\phi = \frac{\pi}{4}$ with the $z$-axis).
• $\theta$: From $0$ to $2\pi$ (full rotation around the $z$-axis).

Step 5: Triple integral setup

The volume is given by: $V = \int_0^{2\pi} \int_0^{\frac{\pi}{4}} \int_{\frac{\rho}{\sqrt{2}}}^a \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.$

Would you like to proceed with the full evaluation of this integral?