# calc

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.

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1. 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

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posted by Damon

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