Estimating Volume with Midpoint Rule for Rotated Region

[wildcattennis]

Homework Statement

If the region shown in the figure is rotated about the -axis to
form a solid, use the Midpoint Rule with to estimate the
volume of the solid.

Homework Equations

Volume formula for cylindrical shells:
V=∫[a,b] 2∏xf(x)dx
midpoint rule:
Δx[f((x0+x1)/2) + f((x1+x2)/2) + ... + f((xn-1+xn)/2)]

The Attempt at a Solution

I started out by finding the area of the figure by using the midpoint rule to approximate the area between the curves.
mpx1= 1
mpx2= 3
mpx3= 5
mpx4= 7
mpx5= 9

Δx= 2, since (b-a)/n is (10-0)/5=2
so midpoint formula used to find the area between the curves using the graph:
2[(f(1)-g(1)) + (f(3)-g(3)) + (f(5)-g(5)) + (f(7)-g(7)) + (f(9)-g(9))]=
2[(4-2) + (5-1) + (4-1) + (4-2) + (4-2)]=
2[2+4+3+2+2]= 26
I then took this number and plugged it in for the f(x) in the volume formula:
v=∫[0,10] 2∏x(26)dx =
2∏∫[0,10] 26xdx = 2∏[13x^2][0,10]=
2∏[13(10)^2] = 2600∏

this is the answer i came up with and i want to make sure it is correct. thank you!

 

[Mark44]

Homework Statement

If the region shown in the figure is rotated about the -axis to
form a solid, use the Midpoint Rule with to estimate the
volume of the solid.

Homework Equations

Volume formula for cylindrical shells:
V=∫[a,b] 2∏xf(x)dx
midpoint rule:
Δx[f((x0+x1)/2) + f((x1+x2)/2) + ... + f((xn-1+xn)/2)]

The Attempt at a Solution

I started out by finding the area of the figure by using the midpoint rule to approximate the area between the curves.
mpx1= 1
mpx2= 3
mpx3= 5
mpx4= 7
mpx5= 9

Δx= 2, since (b-a)/n is (10-0)/5=2
so midpoint formula used to find the area between the curves using the graph:
2[(f(1)-g(1)) + (f(3)-g(3)) + (f(5)-g(5)) + (f(7)-g(7)) + (f(9)-g(9))]=
2[(4-2) + (5-1) + (4-1) + (4-2) + (4-2)]=
2[2+4+3+2+2]= 26
I then took this number and plugged it in for the f(x) in the volume formula:
v=∫[0,10] 2∏x(26)dx =
2∏∫[0,10] 26xdx = 2∏[13x^2][0,10]=
2∏[13(10)^2] = 2600∏

this is the answer i came up with and i want to make sure it is correct. thank you!

I think your answer is too large by an order of magnitude. Where you went wrong was in integrating. What you should be doing is adding five numbers, where each number represents the approximate volume of one shell.

For example, when the first area element is revolved around the y-axis, the volume swept out is
$2\pi (1 \cdot 2 \cdot 2)$

Inside the parentheses, the 1 represents the radius and the first 2 represents the distance between the upper curve and lower curve. The second 2 represents Δx. From your drawing, this part of the region is about 4 sq. units.

You need to do this for each of the five regions. In my work I get $244\pi$.

 

[wildcattennis]

I think your answer is too large by an order of magnitude. Where you went wrong was in integrating. What you should be doing is adding five numbers, where each number represents the approximate volume of one shell.

For example, when the first area element is revolved around the y-axis, the volume swept out is
$2\pi (1 \cdot 2 \cdot 2)$

Inside the parentheses, the 1 represents the radius and the first 2 represents the distance between the upper curve and lower curve. The second 2 represents Δx. From your drawing, this part of the region is about 4 sq. units.

You need to do this for each of the five regions. In my work I get $244\pi$.

thank you.

 

[Mark44]

Sure, you're welcome!