Double Integrals - interpreting what are my boundaries?

[rocomath]

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).

 

[rootX]

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

 

[HallsofIvy]

If the planes form the "top" and "bottom" you can do this in either of two ways:
The integral
$$\int\int \int_{f(x,y)}^{g(x,y)} dzdydx= \int\int (g(x,y)- f(x,y))dydx$$
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.

 

[rocomath]

If the planes form the "top" and "bottom" you can do this in either of two ways:
The integral
$$\int\int \int_{f(x,y)}^{g(x,y)} dzdydx= \int\int (g(x,y)- f(x,y))dydx$$
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

$$-2\int_0^1\int_0^{1-x^2}(x+y-2)dydy$$

Can I not take advantage of symmetry? B/c everywhere I look has it as ... $$\int_{-1}^1\int_{x^2-1}^{1-x^2}(2-x-y)dydx$$ or am I just messing up my algebra?

 

[rocomath]

I meant -4

 

[HallsofIvy]

Have you calculated those two integrals to see if they are different?

 

[rocomath]

Have you calculated those two integrals to see if they are different?

Uh huh ...

$$-4\int_0^1\int_0^{1-x^2}(x+y-2)dydx$$

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

$$-4\left(\frac 1 2-\frac 1 4+\frac{1-\frac 2 3+\frac 1 5}{2}-2+\frac 2 3\right)=\frac{49}{15}$$

Answer = 64/3

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