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.

Question

a) squares
b) rectangles of height 1

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

help with at least setup?

Answer 1

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