The base of a three-dimensional figure is bound by the line y = 6 - 2x on the interval [-1, 2]. Vertical cross sections that are perpendicular to the x-axis are rectangles with height equal to 2. Find the volume of the figure.

The base of a three-dimensional figure is bound by the line x = -2y - 2 on the interval [-4, -1]. Vertical cross sections that are perpendicular to the y-axis are squares. Find the volume of the figure.

The base of a three-dimensional figure is bound by the line x = sqrt(y)+2 on the interval [1, 9]. Vertical cross sections that are perpendicular to the y-axis are squares. Find the volume of the figure.

How do I solve these problems? I don't need the answers, just guidance on how to do it.

draw a sketch

If the interval means from x = -1 to x = +2, then it is also from y = 2 to y = 8

As I read it then this base is a triangle with corners at
(-1,8) (2,2) and (-1,2)

The volume of the figure would then be 2 times the area of the base
Area of base = (1/2) (3)(6) = 9
so volume = 18

To find the volume of a three-dimensional figure, you need to integrate the cross-sectional area along the given axis over the interval of the base. Here's how you can approach each problem:

1. The base is bound by the line y = 6 - 2x on the interval [-1, 2], and the vertical cross sections are rectangles with height 2.
- Start by visualizing the region bounded by the line y = 6 - 2x and the x-axis on the given interval. This creates a trapezoidal shape.
- To find the length of each cross section along the x-axis, you need to determine the width of the trapezoid at each x-value within the interval.
- You can use the formula for the area of a trapezoid: A = (base1 + base2) * height / 2. In this case, base1 is the y-value of the function and base2 is the x-axis. The height is given as 2.
- Set up the integral: ∫[a, b] A(x) dx, where A(x) is the area of the cross section at x.
- Evaluate the integral and find the volume.

2. The base is bound by the line x = -2y - 2 on the interval [-4, -1], and the vertical cross sections are squares.
- Visualize the region bounded by the line x = -2y - 2 and the y-axis on the given interval. This creates a triangular shape.
- Determine the side length of each cross section along the y-axis. In this case, it is the distance between the line and the y-axis at each y-value.
- Set up the integral: ∫[c, d] A(y) dy, where A(y) is the area of the cross section at y.
- Evaluate the integral and find the volume.

3. The base is bound by the line x = sqrt(y) + 2 on the interval [1, 9], and the vertical cross sections are squares.
- Visualize the region bounded by the line x = sqrt(y) + 2 and the y-axis on the given interval. This creates a curved shape.
- Determine the side length of each cross section along the y-axis. In this case, it is the distance between the curve and the y-axis at each y-value.
- Set up the integral: ∫[e, f] A(y) dy, where A(y) is the area of the cross section at y.
- Evaluate the integral and find the volume.

Remember to choose the correct limits of integration based on the interval of the base, and ensure that the integrals represent the area of each cross section in the appropriate axis.