Can someone tell me if im doing this right?? mass using double integrals

[een]

1. The problem statement, all variables and given/known data
A plate occupies the triangular region with vertices (0,0), (0,1), (2,1). The mass density is given by the function \lambda(x,y) = x+y. Find the mass of the plate,


2. Relevant equations



3. The attempt at a solution
This is what I have:
M = \int0-2\intx-1 k(x+y)dydx
k\int0-2 [y^2/2]x-1
=k \int0-2(x^2/2)-1/2dx
=k[x^3/6]0-2 = 8k/6 = 4k/3

[Dick]

You got the right answer but that seems to be more or less by accident. Just to get started why do you think the lower limit of the y integration should be x? And why do you think the integral of (x+y) dy is y^2/2? What happened to the x?

[een]

i used x as the lower limit because it was the lower bound on the graph and in (x+y)dy i thought since it would with respect to y that the x would be zero.....is that right?

[Dick]

i used x as the lower limit because it was the lower bound on the graph and in (x+y)dy i thought since it would with respect to y that the x would be zero.....is that right?

I don't think the lower limit is x. The lower limit lies on the line connecting (0,0) and (2,1). That's not y=x. It's a different line. And no, the integral of x dy isn't zero! When you are integrating dy, treat x as an constant. This isn't differentiating. What's the integral of c dy where c is a constant?

[een]

the integral of c dy would be cy so (x+y)dy would be xy + y^2/2 right? and i dont understand what the lower limit would be. i thought that the lower limit of dy would be x and the upper limit would be 1. and for x the lower limit would be zero and the upper limit would be 2....

[Dick]

Integral of (x+y)dy is xy+y^2/2, yes. You've got that part. For a given value of x, your upper limit is 1, because y=1 is the line connecting the vertices (0,1) and (2,1). Your lower y value must be on the line connecting (0,0) and (2,1). What's the equation of that line in the form y=ax+b? Go back to when you were doing point-slope type problems.

[een]

this is what i got... for x- the lower limit is 0 and the upper limit is 2. for y- the lower limit is 2x and the upper limit is 1..... is that right?

[Dick]

this is what i got... for x- the lower limit is 0 and the upper limit is 2. for y- the lower limit is 2x and the upper limit is 1..... is that right?

Think about that. If the lower limit is 2x then if x=2, the lower limit is 4. That doesn't sound right, does it?

[een]

oh yea its 1/2x...i just typed it wrong... the rest of it is right tho?

[Dick]

oh yea its 1/2x...i just typed it wrong... the rest of it is right tho?

Yes, the lower limit is (1/2)*x or just x/2 (use more parentheses - 1/2x could also mean 1/(2x)), the rest of it's right. Did you get the mass right?

[een]

yea i got it right now.... thanks!

[Dick]

Good job.