Let’s set up the integrals for each problem step by step.

Problem 9 (Cone-shaped tank):

We are tasked with finding the work required to pump water out of a cone-shaped tank. The radius of the base is 4 meters, and the height is 7 meters. The tank is oriented with its pointed end at the bottom.

Step-by-step process:

1. Volume of water at a given height: At a height $y$ above the cone’s vertex, the radius of a horizontal slice is proportional to $y$. Using similar triangles, the radius $r(y)$ at height $y$ is given by: $r(y) = \frac{4}{7}y$ The area of a cross-section at height $y$ is $\pi r(y)^2 = \pi \left( \frac{4}{7}y \right)^2$.

2. Work for a small slice: A thin horizontal slice of water at height $y$ has volume $dV = \pi \left( \frac{4}{7}y \right)^2 dy$, and the mass of the water in this slice is $\rho dV$, where $\rho = 1000 \ \text{kg/m}^3$ is the density of water.

   The weight of the slice is: $\text{Weight} = 1000 \cdot 9.8 \cdot \pi \left( \frac{4}{7}y \right)^2 dy$

3. Work to pump out the slice: The work to pump the water from height $y$ to the top of the tank (7 meters) is the weight of the slice multiplied by the distance it needs to be lifted: $\text{Work} = 1000 \cdot 9.8 \cdot \pi \left( \frac{4}{7}y \right)^2 (7 - y) dy$

4. Set up the integral: The total work is found by integrating this expression from $y = 0$ to $y = 7$: $W = 1000 \cdot 9.8 \cdot \pi \int_0^7 \left( \frac{4}{7}y \right)^2 (7 - y) dy$ Simplify the constants: $W = 1000 \cdot 9.8 \cdot \pi \cdot \frac{16}{49} \int_0^7 y^2 (7 - y) dy$ This is the integral that represents the work required.

Problem 10 (Spherical tank):

Here, we have a spherical tank with an 8-meter radius. The pump is located 6 meters above the top of the tank.

Step-by-step process:

1. Coordinate system and volume of a slice: Let’s place the origin of the coordinate system at the center of the sphere. The equation of the sphere is: $x^2 + y^2 = 8^2$ For a thin horizontal slice of water at height $y$, the radius of the slice is $r(y) = \sqrt{64 - y^2}$, and the volume of the slice is: $dV = \pi r(y)^2 dy = \pi (64 - y^2) dy$

2. Work for a small slice: The weight of the water in the slice is: $\text{Weight} = 1000 \cdot 9.8 \cdot \pi (64 - y^2) dy$

3. Distance to pump the water: The water at height $y$ must be pumped to a point 6 meters above the top of the tank, i.e., a distance of $14 - y$ meters (since the top of the tank is at $y = 8$).

4. Set up the integral: The total work is given by integrating from $y = -8$ to $y = 8$: $W = 1000 \cdot 9.8 \cdot \pi \int_{-8}^8 (64 - y^2)(14 - y) dy$ This is the integral that represents the work required to pump all the water out of the spherical tank.

Would you like further details or explanations about any part of the process?

Here are 5 related questions you might consider:

1. How would the work integral change if the density of the liquid were different from water?
2. How can the volume of the cone or sphere be computed directly, without using integrals?
3. How would the integral change if the tanks had different orientations?
4. What would happen if the pump height in Problem 10 were reduced to the top of the sphere?
5. How can we numerically compute the integrals if required?

Tip: When solving work problems involving tanks, always remember that the limits of integration correspond to the height of the water, and the integrand typically involves the volume and weight of the water slice.