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

1. Apr 4, 2012

s3a

1. The problem statement, all variables and given/known data
Find the volume of the region bounded by the planes 7x + 6y + 8z = 9, y = x, x = 0, z = 0.

2. Relevant equations
Multiple integration.

3. 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!

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103.7 KB
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382
2. Apr 5, 2012

HallsofIvy

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.

3. Apr 5, 2012

s3a

My latest triple integral (in the image I just attached) is wrong. Did I misinterpret what you said?

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18.1 KB
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224
4. Apr 6, 2012

s3a

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?

5. Apr 7, 2012

hawaiifiver

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: Apr 7, 2012
6. Apr 7, 2012