Double Integrals - interpreting what are my boundaries?

  • Thread starter rocomath
  • Start date
  • #1
1,752
1
Hmm ...

The solid enclosed by the parabolic cylinders y=1-x^2, y=x^2-1 and the planes x+y+z=2, 2x+2y-z+10=0 by subtracting two volumes.

Ok, so I know what my limits should be from the parabolic cylinders, but how was I supposed to know that the 2x+2y-z+10=0 plane doesn't play a role in my integral? When I was initially doing this problem, I thought I had to subtract the top plane from the bottom plane (lol I know, but I was just trying things).
 
Last edited:

Answers and Replies

  • #2
378
2
If you are doing this in 2-d i.e. using dx.dy
the plane only define height (so nothing to do with limits)
Remember: Areas and volumes
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,833
956
If the planes form the "top" and "bottom" you can do this in either of two ways:
The integral
[tex]\int\int \int_{f(x,y)}^{g(x,y)} dzdydx= \int\int (g(x,y)- f(x,y))dydx[/tex]
The integral of dzdxdy is just the integral of the limits for z integrated with respect to x and y.

If, as I said, the planes for the top and bottom, solve their equations for z and ust those for g(x,y) and f(x,y). As for the integration in the xy plane, determine where the parabolas intersect, integrate between those x vaues and for the dy integral, integrate from y= x2-1 to y= 1- x2.
 
  • #4
1,752
1
If the planes form the "top" and "bottom" you can do this in either of two ways:
The integral
[tex]\int\int \int_{f(x,y)}^{g(x,y)} dzdydx= \int\int (g(x,y)- f(x,y))dydx[/tex]
The integral of dzdxdy is just the integral of the limits for z integrated with respect to x and y.

If, as I said, the planes for the top and bottom, solve their equations for z and ust those for g(x,y) and f(x,y). As for the integration in the xy plane, determine where the parabolas intersect, integrate between those x vaues and for the dy integral, integrate from y= x2-1 to y= 1- x2.
Halls, this exercise is in the double-integral section, but I do love triple :)

Question though ... so I have it set up as

[tex]-2\int_0^1\int_0^{1-x^2}(x+y-2)dydy[/tex]

Can I not take advantage of symmetry? B/c everywhere I look has it as ... [tex]\int_{-1}^1\int_{x^2-1}^{1-x^2}(2-x-y)dydx[/tex] or am I just messing up my algebra?
 
  • #5
1,752
1
I meant -4
 
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,833
956
Have you calculated those two integrals to see if they are different?
 
  • #7
1,752
1
Have you calculated those two integrals to see if they are different?
Uh huh ...

[tex]-4\int_0^1\int_0^{1-x^2}(x+y-2)dydx[/tex]

[tex]-4\int_0^1\left[x(1-x^2)+\frac{(1-x^2)^2}{2}-2(1-x^2)\right]dx[/tex]

[tex]-4\left(\frac 1 2-\frac 1 4+\frac{1-\frac 2 3+\frac 1 5}{2}-2+\frac 2 3\right)=\frac{49}{15}[/tex]

Answer = 64/3

I guess I can't take advantage of symmetry on this problem.
 

Related Threads on Double Integrals - interpreting what are my boundaries?

  • Last Post
Replies
7
Views
1K
Replies
0
Views
1K
Replies
3
Views
4K
  • Last Post
Replies
5
Views
1K
Replies
15
Views
4K
Replies
5
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
1K
Replies
6
Views
4K
Top