Finding volume/mass of 3D Solid

  • Thread starter Thread starter dinospamoni
  • Start date Start date
  • Tags Tags
    3d Solid
Click For Summary
SUMMARY

The discussion focuses on calculating the volume and mass of a 3D solid defined by the planes z1 = 2x + 4y + 7 and z2 = 3x + 6y + 8, over a square region in the (x, y) plane with corners (4, 1), (5, 1), (5, 2), and (4, 2). The correct setup for volume is V = ∫∫∫ dz dx dy with limits 2x + 4y + 7 < z < 3x + 6y + 8, and for mass, m = ∫∫∫ 880y dz dx dy. The initial incorrect calculations of volume (8.5) and mass (10333.3) were due to using the wrong density value of 800y instead of the correct 880y.

PREREQUISITES
  • Understanding of triple integrals in calculus
  • Familiarity with the concept of density in physics
  • Knowledge of setting limits of integration for multivariable functions
  • Experience with evaluating integrals in multiple dimensions
NEXT STEPS
  • Review the calculation of triple integrals in multivariable calculus
  • Study the application of density functions in mass calculations
  • Learn about the geometric interpretation of volume under surfaces
  • Practice problems involving integration over defined regions in the (x, y) plane
USEFUL FOR

Students in calculus or physics courses, educators teaching multivariable integration, and anyone needing to calculate volume and mass of solids defined by functions in three-dimensional space.

dinospamoni
Messages
90
Reaction score
0

Homework Statement



The volume shown lies above a square in the (x, y) plane having corners (4, 1), (5, 1), (5, 2), and (4, 2). The remaining two surfaces are planes defined by z1 = 2x + 4y + 7 and z2 = 3x + 6y + 8. The density of this object is 880.y. Determine a) the volume of this object, and b) the mass of this object.

A picture is attached

Homework Equations



I can't use the coding on here, so bear with me:

Volume of a 3D object is given by V=∫∫∫dxdydz

Mass of a 3D object is: m=∫∫∫ρdxdydx
where ρ is the density


The Attempt at a Solution



My setup for the volume is:

∫∫∫dzdxdy
Limits of integration:
2x+4y+7 < z < 3x + 6y + 8.
4<x<5
1<y<2

And for mass:

∫∫∫800y dzdxdy

with the same limits of integration

I've gotten 8.5 and 10333.3 for the volume and mass, respectively, but those are wrong.

Can anyone see what I'm doing wrong?
 

Attachments

  • prob09-06.gif
    prob09-06.gif
    4.2 KB · Views: 579
Physics news on Phys.org
Your work looks correct to me. I get the same answers as you.
 
AH I see what I did wrong. When I calculated the mass I used density = 800y instead of 880y
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K