Use a double integral to find volume of solids ?

[CalleighMay]

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 ;)

 

[Defennder]

Well this is one of those maths methods which are essentially superseded when you learn a more general way of getting the same answer. More specifically when you learn triple integrals and volume, you might not remember this particular method.

But anyway, in this case, you have to make use of a general formula for evaluating volume bounded above by function f(x,y) and below by plane surface R. $$\iint_R f(x,y) dA$$.

In this case, you can easily discern the height function f(x,y) using the equation of the solid. For limits, you should be able to see where lines parallel to the x and y-axis would enter and exit the solid.

 

[CalleighMay]

Thanks for the reply!

I am having problems integrating but i got:

integral from 0 to 2 of the integral from 0 to 2-x of (2-x-y)dydx.

I got: -2(3y-4)/3

How do i integrate correctly? I'm pretty sure that's wrong since I've never integrated with a y before... Thanks =/

 

[Defennder]

Your final answer must be a numerical value since you are given the equation of the solid along with the vertices as actual numbers and not unknown constants. How did you get that answer with y inside? Integrate with respect to y first, treating x with a constant, then evaluate the integral limits by subsituting them for y in the antiderivative. Then perform the intgegration with respect to x.