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

1. Nov 3, 2009

### een

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

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 = $$\int$$0-2$$\int$$x-1 k(x+y)dydx
k$$\int$$0-2 [y^2/2]x-1
=k $$\int$$0-2(x^2/2)-1/2dx
=k[x^3/6]0-2 = 8k/6 = 4k/3

2. Nov 3, 2009

### Dick

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

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?

3. Nov 3, 2009

### een

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

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?

4. Nov 3, 2009

### Dick

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

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?

5. Nov 3, 2009

### een

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

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....

6. Nov 3, 2009

### Dick

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

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.

7. Nov 4, 2009

### een

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

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?

8. Nov 4, 2009

### Dick

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

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?

9. Nov 4, 2009

### een

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

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

10. Nov 4, 2009

### Dick

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

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?

11. Nov 4, 2009

### een

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

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

12. Nov 4, 2009

### Dick

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

Good job.