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.

it is symmetric around the y axis so do the integral from x = 0 to x = 1 and double it

2 int from 0 to 1 of x^2 dx
2 (x^3)/3 = 2/3

length of side of square = 2x^2
area of cross section = 4 x^4
integrate 4 x^4 dx from 0 to 1

now do the same for area of cross section = x^4 sqrt 3

I did the area below the line. For between use (1-x^2) dx etc

What is symmetric around the y-axis? This is confusing because I don't really know how to draw the solid. First it says the solid has the base of R. So does that mean R should be on the x-axis? But how do you find out the length of the side of the square? Is one side of the square solid bounded by the region so that one of the sides of the square faces of the solid supposed to have a length that is from one intersection point of both curves to the other intersection point? Same goes for part c. Thanks

(a) To find the area of region R, you can split it into two parts: the area between the curve y = x^2 and the line y = 1, and the area below the x-axis.

Let's start by finding the area between the curve y = x^2 and the line y = 1. To do this, we need to find the intersection points of these two curves, which will give us the limits of integration.

Setting y = x^2 and y = 1 equal to each other, we get x^2 = 1. Taking the square root of both sides, we find x = ±1.

So the limits of integration for x are -1 and 1.

The limits of integration for y are given by the equations of the curves that bound the region. The lower bound is the x-axis, which is y = 0. The upper bound is the line y = 1.

Therefore, the integral to find the area of R is:

Area = ∫[from -1 to 1]∫[from 0 to 1] dy dx.

Integrating this double integral will give you the area of region R.

(b) To find the volume of the solid with base R and cross-sections perpendicular to the y-axis that are squares, we'll use the method of slicing.

Each square cross-section has a side length equal to the difference between the y-coordinates of the boundary curves at the corresponding y-value. In this case, the boundary curves are y = x^2 and y = 1.

For a given value of y, the left boundary of the square cross-section is at x = -√y, and the right boundary is at x = √y. Therefore, the side length of each square cross-section is (√y - (-√y)) = 2√y.

To find the volume of the solid, we'll integrate the area of each square cross-section from y = 0 to y = 1, since that's the range of y-values that R covers.

So the integral to find the volume is:

Volume = ∫[from 0 to 1] (2√y)^2 dy

Simplifying this integral and evaluating it will give you the volume of the solid.

(c) Similar to part (b), to find the volume of the solid with base R and cross-sections perpendicular to the y-axis that are equilateral triangles, we'll use the method of slicing.

Each equilateral triangle cross-section has a height equal to the difference between the y-coordinates of the boundary curves at the corresponding y-value. In this case, the boundary curves are y = x^2 and y = 1.

For a given value of y, the bottom boundary of the equilateral triangle cross-section is at x = -√y, and the top boundary is at x = √y. Therefore, the height of each equilateral triangle cross-section is (√y - (-√y)) = 2√y.

The area of each equilateral triangle is given by A = (sqrt(3)/4) * s^2, where s is the length of one side. In this case, the side length s is equal to 2√y.

To find the volume of the solid, we'll integrate the area of each equilateral triangle cross-section from y = 0 to y = 1.

So the integral to find the volume is:

Volume = ∫[from 0 to 1] [(sqrt(3)/4) * (2√y)^2] dy

Simplifying this integral and evaluating it will give you the volume of the solid.