Find the volume of the region bounded by the planes (Multiple Integration).

Click For Summary
SUMMARY

The volume of the region bounded by the planes defined by the equations 7x + 6y + 8z = 9, y = x, x = 0, and z = 0 can be calculated using multiple integration techniques. The correct setup for the triple integral is established as $$ \int_0^{\frac{9}{7}} \int_0^{\frac{9-7x}{6}} \frac{9-7x-6y}{8} dy dx $$, which accounts for the limits of integration based on the intersection points of the planes. The initial incorrect limits in the dy integral were identified as a source of error in the calculations. The final volume can be computed accurately using this corrected integral setup.

PREREQUISITES
  • Understanding of multiple integration techniques
  • Familiarity with the concept of bounded regions in three-dimensional space
  • Knowledge of plane equations and their intersections
  • Experience with evaluating triple integrals
NEXT STEPS
  • Study the method of setting up triple integrals for volume calculations
  • Learn about the geometric interpretation of multiple integrals
  • Practice solving similar problems involving bounded regions and planes
  • Explore the use of computational tools like Wolfram Alpha for verifying integral calculations
USEFUL FOR

Students in calculus courses, particularly those focusing on multivariable calculus, as well as educators seeking to clarify the concepts of multiple integration and volume calculations in three-dimensional geometry.

s3a
Messages
828
Reaction score
8

Homework Statement


Find the volume of the region bounded by the planes 7x + 6y + 8z = 9, y = x, x = 0, z = 0.

Homework Equations


Multiple integration.

The Attempt at a Solution


My attempt at a solution is attached. To test, I computed the answer with Wolfram Alpha which yielded an incorrect answer of 243/5488.

Any help in solving this problem would be greatly appreciated!
Thanks in advance!
 

Attachments

  • MyWork.jpg
    MyWork.jpg
    57 KB · Views: 1,390
Physics news on Phys.org
You have decided to integrate with respect to z, then y, then x so look at the projection down to the plane z= 0. When z= 0 the plane 7x+ 6y+ 8z= 9 becomes the line 7x+ 6y= 9 which intersects the x-axis at (9/7, 0) and the y-axis at (0, 3/2).

So x will range from 0 to 9/7 and, for each x, y will range from 0 to (9- 7x)/6. In your attachment, you have the equation y= (9- 7x)/6 but, for some reason, you show the upper limit of the "dy" integral as just x rather than (9- 7x)/6.
 
My latest triple integral (in the image I just attached) is wrong. Did I misinterpret what you said?
 

Attachments

  • TripleIntegral.jpg
    TripleIntegral.jpg
    9.7 KB · Views: 1,035
I re-attempted the problem based on some new stuff I've learned but I still get a wrong answer. (I attached my latest work.) Could you please tell me what's wrong in the setup?
 
I've been reading how to do this, you want to take a double integral of your plane as follows:

$$ \int_0^{\frac{9}{7}} \int_0^{\frac{9-7x}{6}} \frac{9-7x-6y}{8} dy dx $$
 
Last edited:

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
7
Views
4K