Find volume of the solid: under z=xy, above triangle w/ vert

[MarkSA]

Hi,

Q: Find volume under z=xy and above triangle with vertices (1,1) and (9,1) and (1,2)

I'm having troubles figuring out what the limits of integration should be in these problems. I sketched the triangle in question.
I know I need to take the double integral of (xy) with respect to first one varible and then the other. I don't know if the order matters or how to determine which to do first (dx or dy.)

But just from eyeballing the graph, my first guess would be that I need to integrate from y=1 to y=2 and from x=1 to x=9. This isn't right though and I don't understand why. The correct limits are apparently y=1 to y=2 and x=1 to x=17-8y (gotten from the equation of the hypotenuse). I don't understand why the dy limits can be numerical values but the dx limits use the equation of the line. And on another topic, why couldn't the limits be from y=1 to y= -1/8*x +17/8 and from x=1 to x=9? Can you help me understand this?

 

[Deleted member 4993]

Re: Find volume of the solid

Q: Find volume under z=xy and above triangle with vertices (1,1) and (9,1) and (1,2)<<< What's the z-coordinate of these points?

 

[MarkSA]

Re: Find volume of the solid

(1,1) and (9,1) and (1,2)<<< What's the z-coordinate of these points?

(1,1) --> z = 1
(9,1) --> z = 9
(1,2) --> z =2

 

[PAULK]

Re: Find volume of the solid

Hi,

Q: Find volume under z=xy and above triangle with vertices (1,1) and (9,1) and (1,2)

I'm having troubles figuring out what the limits of integration should be in these problems. I sketched the triangle in question.
I know I need to take the double integral of (xy) with respect to first one varible and then the other. I don't know if the order matters or how to determine which to do first (dx or dy.)

But just from eyeballing the graph, my first guess would be that I need to integrate from y=1 to y=2 and from x=1 to x=9. This isn't right though and I don't understand why. The correct limits are apparently y=1 to y=2 and x=1 to x=17-8y (gotten from the equation of the hypotenuse). I don't understand why the dy limits can be numerical values but the dx limits use the equation of the line. And on another topic, why couldn't the limits be from y=1 to y= -1/8*x +17/8 and from x=1 to x=9? Can you help me understand this?

I make it like this:

Your x is surely going from 1 to 9 -- easy.
And your y is going from the lower boundary: y = 1,
to an upper boundary y = the line passing through (1,2) and (9,1)

which I make to be: y = - x/8 + 17/8

So you integrate your z = xy:

{9 {-x/8+17/8
| | (xy) dy dx
}1 }1

-- definitely in that order; first the dy-integration, considering x to be a constant, and substituting those messy boundaries, then x.

HOWEVER, you can also do it your way: Have y go from 1 to 2, and x go from 1 to 17- 8y. You are definitely on the right track; just remember -- if you do the x-integration first, you have variable boundaries in terms of y.

Good luck.

 

[MarkSA]

Re: Find volume of the solid

Thanks Paul, these problems are making more sense now.