find the volume of the solid whose base is bounded by the graphs of y= x+1 and y= (x^2)+1, with the indicated cross sections taken perpendicular to the x-axis.

a) squares
b) rectangles of height 1

the answers are supposed to be
a. 81/10
b. 9/2

help with at least setup?

First, we need to find the points where the two curves y = x + 1 and y = x^2 + 1 intersect. We set them equal to each other:

x + 1 = x^2 + 1
x^2 - x = 0
x(x - 1) = 0

So the two curves intersect at x = 0 and x = 1.

Now we'll set up the volumes for each shape in parts.

a) Squares

The side length of the squares is the difference between the two functions: (x + 1) - (x^2 + 1) = 1 - x^2. So the area of each square is (1 - x^2)^2. We'll integrate from 0 to 1 to find the volume.

V = ∫[(1 - x^2)^2]dx from 0 to 1
= ∫[(1 - 2x^2 + x^4)]dx from 0 to 1
= [x - (2/3)x^3 + (1/5)x^5] from 0 to 1
= (1 - 2/3 + 1/5) - (0)
= (5/5 - 10/15 + 3/15)
= 81/30

b) Rectangles of height 1

The length of the rectangles is the same as the side length of the squares: 1 - x^2. The height is given as 1. We'll again integrate from 0 to 1 to find the volume.

V = ∫[(1 - x^2)]dx from 0 to 1
= ∫(_)dx from 0 to 1
= [x - (1/3)x^3] from 0 to 1
= (1 - 1/3) - (0)
= 2/3

So the volumes are:
a) 81/30
b) 2/3

To find the volume of the solid with cross sections taken perpendicular to the x-axis, we need to integrate the areas of the cross sections along the x-axis.

a) Squares:
The cross sections will be squares with side length equal to the difference between the two given curves, which is (x+1) - ((x^2)+1) = x - (x^2).

To set up the integral, we need to find the limits of integration, which are the x-values at which the two curves intersect. Setting the equations equal to each other:
x + 1 = x^2 + 1
x^2 - x = 0
x(x - 1) = 0

So, x = 0 and x = 1 are the limits of integration.

The integral for the volume is then:
V = ∫[0 to 1] (x - (x^2))^2 dx

b) Rectangles of height 1:
The cross sections will be rectangles with base width equal to the difference between the two given curves, which is (x+1) - ((x^2)+1) = x - (x^2), and height 1.

To set up the integral, again we need to find the limits of integration, which are the x-values at which the two curves intersect. Using the same equations as above, x = 0 and x = 1 are the limits of integration.

The integral for the volume is then:
V = ∫[0 to 1] (x - (x^2)) dx

Now, you can evaluate the integrals using the limits of integration and calculate the volume for both cases.

To find the volume of a solid with cross sections perpendicular to the x-axis, we can use the method of integration.

a) Squares:
First, we need to find the limits of integration by determining the x-values where the graphs of y = x + 1 and y = x^2 + 1 intersect. Set the two equations equal to each other:

x + 1 = x^2 + 1

Rearranging, we get:

x^2 - x = 0

Factoring out x, we have:

x(x - 1) = 0

So, x = 0 or x = 1.

The limits of integration for the base will be from x = 0 to x = 1.

Now, let's consider a cross section where the height is a square. The length of the base of this square will be the difference in y-values of the two curves: (x^2 + 1) - (x + 1) = x^2 - x.

Therefore, the area of this square cross section is (x^2 - x)^2 = x^4 - 2x^3 + x^2.

To find the volume, we integrate the area of the cross section over the interval [0, 1]:

V = ∫[0,1] (x^4 - 2x^3 + x^2) dx

Evaluating this integral will give us the volume of the solid.

b) Rectangles of height 1:
Similar to the previous case, we find the limits of integration by setting the two equations equal to each other:

x + 1 = x^2 + 1

Which gives us x^2 - x = 0.

The limits of integration for the base will be from x = 0 to x = 1.

Now, consider a cross section where the height is a rectangle of height 1. The length of the base of this rectangle will be the difference in y-values of the two curves: (x^2 + 1) - (x + 1) = x^2 - x.

Therefore, the area of this rectangle cross section is (x^2 - x) * 1 = x^2 - x.

To find the volume, we integrate the area of the cross section over the interval [0, 1]:

V = ∫[0,1] (x^2 - x) dx

Evaluating this integral will give us the volume of the solid.