Volume of cross sections using rectangles

  • Thread starter Thread starter mikaloveskero
  • Start date Start date
  • Tags Tags
    Cross Volume
Click For Summary

Homework Help Overview

The problem involves finding the volume of a solid with rectangular cross-sections perpendicular to the x-axis, bounded by the curve y=√x, the x-axis, and the line x=9. The height of the rectangles is defined as half the base length.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the setup of the problem, questioning the clarity of the original poster's description and the correctness of their calculations. There are attempts to clarify the relationships between the dimensions of the cross-sections and the area under the curve.

Discussion Status

Some participants have pointed out potential errors in the original calculations and have sought clarification on the geometric setup. Multiple interpretations of the problem are being explored, with participants noting discrepancies in the derived volume results.

Contextual Notes

There is an indication of confusion regarding the definitions of the dimensions of the cross-sections and the integration process. The original poster has acknowledged errors in their work but has not resolved the discrepancies in their answers.

mikaloveskero
Messages
6
Reaction score
0

Homework Statement



The area bounded by y=√x, the x-axis and the line x=9 and is perpendicular to the x axis. find the volume of the cross section using rectangle with the h=1/2b

Homework Equations





The Attempt at a Solution


I just want to know if I'm correct or on the right track. if not please correct me

A=HB
A=(1/2B)B =3/2B
So i take the integral of 3/2B from 0 to 9
V=3/2 integral 0 to 9 B
V=3B^2/4 from 0 to 9
V=243/4
 
Physics news on Phys.org
mikaloveskero said:

Homework Statement



The area bounded by y=√x, the x-axis and the line x=9 and is perpendicular to the x axis. find the volume of the cross section using rectangle with the h=1/2b

Homework Equations





The Attempt at a Solution


I just want to know if I'm correct or on the right track. if not please correct me

A=HB
A=(1/2B)B =3/2B
So i take the integral of 3/2B from 0 to 9
V=3/2 integral 0 to 9 B
V=3B^2/4 from 0 to 9
V=243/4

You haven't stated the problem clearly enough to be sure what it actually is, but I'm pretty certain your solution is wrong anyway.

Does the solid you are trying to describe sit on the described area with rectangular cross-sections perpendicular to the xy plane? Are the bases of the rectangular cross-sections in the xy plane perpendicular to the x-axis?
 
LCKurtz said:
You haven't stated the problem clearly enough to be sure what it actually is, but I'm pretty certain your solution is wrong anyway.

Does the solid you are trying to describe sit on the described area with rectangular cross-sections perpendicular to the xy plane? Are the bases of the rectangular cross-sections in the xy plane perpendicular to the x-axis?

Yes, it does sit on the area with rectangular cross sections perpendicular to the x and y plane and the bases of the rectangular cross section are perpendicular to the x axis. I just need to find the volume using the given information.

Also i realize an error in my work :
A=HB
A=(1/2B)B =3/2√ x
So i take the integral of 3/2√x from 0 to 9
V=3/2 integral 0 to 9 x^3/2/(3/2)
V=6x^3/2/(6) from 0 to 9
V=162/6

or

The base is going to be sqrt(x). Therefore the height is 1/2sqrt(x), and the area is 1/2 x. The volume will be the integral from 0 to 9 of 1/2 x, which is 81/4.

i get two different answers
 
Last edited:
mikaloveskero said:
Yes, it does sit on the area with rectangular cross sections perpendicular to the x and y plane and the bases of the rectangular cross section are perpendicular to the x axis. I just need to find the volume using the given information.

Also i realize an error in my work :
A=HB
A=(1/2B)B =3/2√ x
So i take the integral of 3/2√x from 0 to 9
V=3/2 integral 0 to 9 x^3/2/(3/2)
V=6x^3/2/(6) from 0 to 9
V=162/6

or

The base is going to be sqrt(x). Therefore the height is 1/2sqrt(x), and the area is 1/2 x. The volume will be the integral from 0 to 9 of 1/2 x, which is 81/4.

i get two different answers

That's because (1/2)B*B doesn't equal (3/2)B.
 

Similar threads

Good approximation - multivariable calculus
  • · Replies 9 ·
Replies
9
Views
2K
Rectangle volume using cross sections
  • · Replies 2 ·
Replies
2
Views
2K
Volume between a region (above x-axis and below parabola) and a surface
  • · Replies 4 ·
Replies
4
Views
2K
Volume of Static Solid Using Cross-Sectional Area (Integration)
  • · Replies 2 ·
Replies
2
Views
1K
Finding Volume of Solid with Rectangle Cross-Sections | Geometry Homework
  • · Replies 1 ·
Replies
1
Views
1K
Use binomial theorem to find the complex number
  • · Replies 2 ·
Replies
2
Views
2K
Triple Integral To Find Volume Between Cylinder And Sphere
  • · Replies 2 ·
Replies
2
Views
2K
Calculating the volume of the solid in this graph
  • · Replies 4 ·
Replies
4
Views
2K
The volume of a "spherical cap" using triple integrals
  • · Replies 9 ·
Replies
9
Views
2K
Volume of cross sections using isosceles right triangle
  • · Replies 2 ·
Replies
2
Views
13K