R is the region in the plane bounded below by the curve y=x^2 and above by the line y=1.

(a) Set up and evaluate an integral that gives the area of R.

(b) A solid has base R and the cross-sections of the solid perpendicular to the y-axis are squares. Find the volume of the solid.

(c) A solid has base R and the cross-sections of the solid perpendicular to the y-axis are equilateral triangles. Find the volume of the solid.

(a) The y = x^2 curve crosses y = 1 at x = 1. For the area in question, compute the integral of x^2 from x = 0 to 1, and subtract it from 1.

(b)Integrate along the y axis, with differential slab volume x^2 dy = y dy, from y=0 to y=1

(c) Proceed similary to (b), integrating along y from 0 to 1, but with equilateral triangle slabs. Slab volume must be expressed in terms of x

Why would it only be the integral of x^2 and not the integral of 1-x^2 from x=-1 to x=1? So wouldn't that equal an area of 4/3? And secondly, how would you sketch these solids? when you say integrate along the y-axis do you mean to say that the curves have to be shifted as well? And why are all the integrals from x=0 to x=1?

Find the volume, V , of the solid obtained

by rotating the region bounded by the graphs
of
x = y2, x = squareroot(y)
about the line x = −1.

To solve these problems, we need to find the limits of integration and set up the appropriate integral based on the given information.

(a) To find the area of region R, we need to integrate the difference between the upper and lower functions with respect to x over the limits of the region.

The upper function is y = 1 (the line), and the lower function is y = x^2 (the curve). To find the limits of integration, we set the two functions equal to each other and solve for x:

x^2 = 1
x = ±1

So the limits of integration for x are -1 and 1. To set up the integral, we integrate the difference between the upper and lower functions:

Area = ∫[from -1 to 1] (1 - x^2) dx

Evaluating this integral will give you the area of region R.

(b) To find the volume of the solid with square cross-sections, we need to integrate the area of each cross-section perpendicular to the y-axis over the limits of the region.

Since the cross-sections are squares, the area of each cross-section is given by the side length squared. The side length of the square at any given y-value is the difference between the upper and lower x-values for that y.

To find the limits of integration, we look at the region R. The curve y = x^2 intersects the line y = 1 at x = -1 and x = 1, so the limits of integration for y are 0 (the y-value at the bottom of region R) and 1.

Therefore, the integral for the volume of the solid is:

Volume = ∫[from 0 to 1] [(upper x-value - lower x-value)^2] dy

(c) To find the volume of the solid with equilateral triangle cross-sections, we need to integrate the area of each cross-section perpendicular to the y-axis over the limits of the region.

The area of each equilateral triangle cross-section can be expressed in terms of its height and base length. For an equilateral triangle, the height is given as the square root of 3 divided by 2 times the side length, and the base length is equal to the side length.

To find the limits of integration, we again look at the region R. The curve y = x^2 intersects the line y = 1 at x = -1 and x = 1, so the limits of integration for y are 0 (the y-value at the bottom of region R) and 1.

Therefore, the integral for the volume of the solid is:

Volume = ∫[from 0 to 1] [(base length * height)] dy

Where the base length can be expressed as (upper x-value - lower x-value), and the height can be expressed as (√3/2) times the base length.

Evaluating this integral will give you the volume of the solid with equilateral triangle cross-sections.