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

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Homework Help Overview

The problem involves finding the volume of a region bounded by the planes defined by the equation 7x + 6y + 8z = 9, along with the constraints y = x, x = 0, and z = 0. The subject area pertains to multiple integration techniques in calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the triple integral for the volume calculation, with some questioning the limits of integration and the interpretation of the bounding planes. There are attempts to clarify the projection onto the z=0 plane and the relationships between the variables.

Discussion Status

There is ongoing exploration of the correct setup for the integral, with some participants providing guidance on the limits of integration. Multiple interpretations of the problem are being considered, and participants are sharing their attempts and corrections without reaching a consensus.

Contextual Notes

Participants mention discrepancies in their calculations and the need to verify the limits of integration based on the geometric interpretation of the planes involved. There is also reference to external computational tools yielding conflicting results.

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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: 1,401
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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: 1,044
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:

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