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

[s3a]

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!

 

[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.

 

[s3a]

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

 

[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?

 

[hawaiifiver]

I've been reading how to do this, you want to take a double integral of your plane as follows:

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

 

[s3a]

I learned more about it and drew it properly and (sucessfully) got the following if anyone cares:

http://www.wolframalpha.com/input/?...y)/8+dy+from+x+to+(9-7x)/6)+dx+from+0+to+9/13

Thanks for the help.