Use a double integral to find volume of solids ?

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CalleighMay

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Hey guys! I have been on the forum for about a week or so and have compiled a lot of information and techniques to help me understand calculus, so i really appreciate everyone's help!

I am a soon-to-be freshman in college and am taking a summer class, calculus II (took calc I in HS). This is our last week of class after our final exam so my professor is taking this time to give us a preview of what we will be learning in the fall semester in Calc III (since this is the same professor). Every Tuesday class our professor gives us a few problems from future sections and asks us to "see what we can come up with" and to work together to find solutions. The following Tuesday he asks us to discuss the problems as a class, seeing which ones of us know our stuff =P

Basically, i want to ask you guys what you think about these problems as i do them along before i have my discussion. I really want to make a lasting impression on my professor by "knowing my stuff" -to show him i can do it! All's i need is a little help! Would you guys mind giving me some help?

We are using the textbook Calculus 8th edition by Larson, Hostetler and Edwards and the problems come from the book.

The problem is on pg 998 in chapter 14.2 in the text, number 26. It reads:

Use a double integral to find the volume of the indicated solid:
And it gives a picture of a solid with the vertices's: (0,0,0), 2,0,0), (0,2,0) and (0,0,2).
The solid is given the equation: x+y+z=2

I looked at similar problems in the same section and came up with a few ideas as to how to get started. This is what i came up with.
Making a sketch of the side on the xy-plane, i have the line: y=-x+2
Then the integral from 0 to 2 of the integral of 0 to (-x+2) of (this is where i get confused)

I get confused from here...

Any help would be greatly appreciated. Thanks guys ;)
 
You are on the right track.

What you have is a tetrahedron.

Solve the equation for z: \(\displaystyle z=2-x-y\)

\(\displaystyle \int_{0}^{2}\int_{0}^{2-x}(2-x-y)dydx\)

Now integrate.
 
K so saw how you solved for z in the equation x+y+z=2.

While doing that integral, i'm a bit confused as to how to integrate with the y, but here goes nothing:
I got: x(2-y) - x^2/2

When plugging the constraints in,
(2-x)(2-y) - (2-x)^2/2

When taking the integral of this:
-x(x^2-3xy+12(y-1)) / 6
When plugging in the constraints:
-2(2^2-32y+12(y-1)) / 6
which i got:
-2(3y-4) / 3

since this is my maybe, second time doing this "double integral" i'm sure it's wrong. I'm confused as to what i do with the y too.

Thanks for your help!!!! :D
 
You tell us. Did you do it correctly? Did you make any mistakes? Can you verify your result? At this level of mathematics, total dependence is not a good plan.
 
i honestly have no clue if what i have done is correct :( I haven't been taught this stuff in class and i don't have a lot of experience with these problems...

That's why i came here to get your help because i KNOW you know how to do it. :)
 
CalleighMay said:
i honestly have no clue if what i have done is correct :( I haven't been taught this stuff in class.... i KNOW you know how to do it. :)
Click to expand...
Yes, we do know. However, to teach it to you would require many hours over days or weeks. This is not something we can do within this environment, especially when you appear unwilling to do anything more than flip through your book looking for an exact-match example to copy.

To repeat once again: While the volunteers here can help students work through specific questions regarding specific exercises or topics, this assistance requires that the student have at least a basic grasp of the underlying material (something you have clearly stated is non-existant) and that the student be willing to contribute to the process (something you have yet to evidence). We can not provide the hours of classroom instruction necessary to provide that foundational underpinning.

I apologize for the confusion, and trust that, if you still do not understand what is being said above, you will invest a few moments to study it and at least attempt to figure it out before replying again.

Eliz.
 
I am only concerned about my answer because for a few reasons..
1. I would have to assume that the work i did is correct because i used a TI-89 that has integration capabilities.
2. I would also assume it's wrong because the answer is not a numerical value, which i believe i should have attained by this integration. I do not know how to integrate 2-x-y by hand.

Do i make it:
2x - 1/2x^2 - 1/2y^2 ?
 
\(\displaystyle \int_{0}^{2}\int_{0}^{2-x}(2-x-y)dydx\)

We first integrate w.r.t y keeping x as a constant.

Then we get \(\displaystyle 2y-xy-\frac{1}{2}y^{2}\).

Using the limits of integration, 0 to 2-x, on y we get \(\displaystyle \frac{(x-2)^{2}}{2}\)

Now, integrate that w.r.t x and we get: \(\displaystyle \frac{(x-2)^{3}}{6}\)

Use the limits for x, 0 to 2, and we get \(\displaystyle \frac{4}{3}\)

That is the solution.
 
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