# need help getting a triple integral problem started

• Jul 27th 2013, 05:12 PM
mrb165
need help getting a triple integral problem started
Find the volume of the region bounded by the paraboloid z = x^2 + y^2 + 4 and the planes x = 0, y = 0, z = 0, x + y = 1

So I believe that the limits are:

z : 0 to x^2 + y^2 + 4

y : 0 to 1-x

x : 0 to.... here's where I'm lost, if I set x = 1-y I wouldn't be able to solve this integral, right?

Also I don't understand what I'm supposed to be integrating, is it just
x^2 + y^2 - z + 4 because it asks for the volume of the paraboloid?
• Jul 27th 2013, 05:27 PM
chiro
Re: need help getting a triple integral problem started
Hey mrb165.

If the region you are integrating is not smooth (but continuous), then you will need to split up the integral into multiple parts. Since you have a paraboloid and a plane, there is a good chance of this happening.

To start you off in the right direction you should find where the x+y=1 plane intersections the paraboloid and look at partitioning the integral based on where this intersection is.

The analogy is that if you integrate say |x|dx over [-5,5] then you need to calculate [-5,0] and [0,5] when you are integrating.
• Jul 27th 2013, 05:31 PM
HallsofIvy
Re: need help getting a triple integral problem started
Start by drawing a picture. If you are really good draw a three dimensional picture, but, fortunately for me, that isn't really necessary. Yes, you are right that for each (x, y), z goes from 0 (because we are told that z= 0 is one boundary) up to z= x^2+ y^2+ 4.

Now, we are down to the limits on the x and y integrals so we are in the xy- plane and that is easy to draw. One boundary is, of course, x= 0, another y= 0, and the third is the line x+ y= 1. That is, of course, the line from (0, 1) to (1, 0). For each x, y goes from 0 up to 1- x. Then, of course, with no further restriction on x (such as "for each x" or "for each (x, y)", x must go from 0 to 1. In order to find the integral, you want to integrate "dV= dzdydz"; $\displaystyle \int\int\int dV= V$.

So you want to integrate $\displaystyle int_0^1\int_0^{1- x}\int_0^{1- x}\int_0^{x^2+ y^2+ 4} dzdydx$. I can see where your confusion is, though- look what you get after doing that first integral, $\displaystyle \int_0^{x^2+ y^2+ 4} dz$.

Other ways to do this: once you are in the xy-plane, you could also argue that, for each y x goes from 0 to 1- y and then ygoes from 0 to 1. That would give
$\displaystyle \int_0^1\int_0^{1- y}\int_0^{x^2+ y^2+ 4} dzdxdy$
• Jul 27th 2013, 06:35 PM
mrb165
Re: need help getting a triple integral problem started
ok, here's how I set it up....

1 1-x x²+y²+4
∫ ∫ ∫ dzdydx
0 0 0

then I got this....

1 1-x
∫ ∫ [x^2 + y^3 +4]dydx
0 0

and then I kept integrating and now I'm stuck at this...

1
∫ ((1-x)^3)/3 + 4 -4x dx
0

I have no idea how to integrate this now, I'm not sure I even did any of this right...
• Jul 27th 2013, 07:03 PM
Prove It
Re: need help getting a triple integral problem started
Quote:

Originally Posted by mrb165
ok, here's how I set it up....

1 1-x x²+y²+4
∫ ∫ ∫ dzdydx
0 0 0

then I got this....

1 1-x
∫ ∫ [x^2 + y^2 +4]dydx
0 0

and then I kept integrating and now I'm stuck at this...

1
∫ ((1-x)^3)/3 + 4 -4x dx
0

I have no idea how to integrate this now, I'm not sure I even did any of this right...

Is this a typo or is this what you actually got?
• Jul 27th 2013, 07:16 PM
mrb165
Re: need help getting a triple integral problem started
Quote:

Originally Posted by Prove It
Is this a typo or is this what you actually got?

yeah, thats what I got after the first integration, then it would become:

1 1-x
∫ [x^2 + (y^3)/3 +4y] dx
0 0

right? (1-x and 0 should be to the right side)
• Jul 27th 2013, 07:40 PM
mrb165
Re: need help getting a triple integral problem started
ok hold on I forgot you add the variable to constants now I have:

1
∫ x^2(1-x) + (1-x)^3/3 + 4(1-x) dx
0

is this right so far? I'm not sure if I know how to integrate this
• Jul 27th 2013, 07:55 PM
Prove It
Re: need help getting a triple integral problem started
Quote:

Originally Posted by mrb165
yeah, thats what I got after the first integration, then it would become:

1 1-x
∫ [x^2 + (y^3)/3 +4y] dx
0 0

right? (1-x and 0 should be to the right side)

\displaystyle \displaystyle \begin{align*} \int_0^{x^2 + y^2 + 4}{dz} = x^2 + y^2 + 4 \end{align*}, NOT \displaystyle \displaystyle \begin{align*} x^2 + y^3 + 4 \end{align*}...
• Jul 27th 2013, 08:31 PM
mrb165
Re: need help getting a triple integral problem started
Attachment 28910

ok, just figured out you can post images... anyway I got -4/3 for the volume so I think it's wrong, not sure where I went wrong though...