To find the volume of the solid that lies within the sphere $x^2 + y^2 + z^2 = 25$, above the $xy$-plane, and outside the cone $z = 7\sqrt{x^2 + y^2}$, we can approach this problem using spherical coordinates.

Step 1: Set up the equations in spherical coordinates

In spherical coordinates:

• $x = \rho \sin\theta \cos\phi$
• $y = \rho \sin\theta \sin\phi$
• $z = \rho \cos\theta$

where:

• $\rho$ is the radial distance from the origin,
• $\theta$ is the angle from the positive $z$-axis (polar angle),
• $\phi$ is the azimuthal angle in the $xy$-plane.

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

Step 2: Convert the sphere and cone equations

1. Sphere: $x^2 + y^2 + z^2 = 25 \Rightarrow \rho^2 = 25 \Rightarrow \rho = 5$.
2. Cone: $z = 7\sqrt{x^2 + y^2}$.

For the cone, substitute $x^2 + y^2 = \rho^2 \sin^2 \theta$ and $z = \rho \cos \theta$: $\rho \cos \theta = 7 \rho \sin \theta$ Dividing both sides by $\rho$ (assuming $\rho \neq 0$): $\cos \theta = 7 \sin \theta$ or $\cot \theta = 7 \Rightarrow \theta = \arctan\left(\frac{1}{7}\right)$

This angle $\theta = \arctan\left(\frac{1}{7}\right)$ represents the boundary of the cone.

Step 3: Set up the integral

We want the volume above the $xy$-plane (so $\theta$ ranges from $0$ to $\arctan\left(\frac{1}{7}\right)$) and outside the cone but within the sphere (so $\rho$ ranges from $0$ to $5$).

The volume $V$ is given by: $V = \int_0^{2\pi} \int_0^{\arctan(1/7)} \int_0^5 \rho^2 \sin \theta \, d\rho \, d\theta \, d\phi$

Step 4: Evaluate the integral

1. Integrate with respect to $\rho$: $\int_0^5 \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_0^5 = \frac{125}{3}$

2. Integrate with respect to $\theta$: $\int_0^{\arctan(1/7)} \sin \theta \, d\theta = \left[ -\cos \theta \right]_0^{\arctan(1/7)} = -\cos\left(\arctan\left(\frac{1}{7}\right)\right) + 1$

To simplify $\cos\left(\arctan\left(\frac{1}{7}\right)\right)$, note that: $\cos\left(\arctan\left(\frac{1}{7}\right)\right) = \frac{7}{\sqrt{1^2 + 7^2}} = \frac{7}{\sqrt{50}} = \frac{7}{5\sqrt{2}} = \frac{\sqrt{2}}{10}$ Thus: $-\cos\left(\arctan\left(\frac{1}{7}\right)\right) + 1 = 1 - \frac{\sqrt{2}}{10}$

3. Integrate with respect to $\phi$: $\int_0^{2\pi} d\phi = 2\pi$

Step 5: Combine results

Putting it all together: $V = \frac{125}{3} \cdot \left(1 - \frac{\sqrt{2}}{10}\right) \cdot 2\pi$ $V = \frac{250\pi}{3} \left(1 - \frac{\sqrt{2}}{10}\right)$

Thus, the volume of the solid is: $V = \frac{250\pi}{3} \left(1 - \frac{\sqrt{2}}{10}\right)$

Would you like further explanation on any part of this solution?

Here are some related questions that might help expand understanding:

1. How does spherical coordinates help in simplifying volume calculations?
2. What is the volume of a cone bounded by a sphere in cylindrical coordinates?
3. How do polar and spherical coordinates differ in applications?
4. Can we calculate the volume directly using Cartesian coordinates for this setup?
5. How would the integral setup differ if the problem asked for volume above the cone and below the sphere?

Tip: When converting to spherical coordinates, it's useful to remember the relationships between $x, y, z$ and $\rho, \theta, \phi$ as they can simplify complex boundaries.