Triple Integral Limits Help

  • Thread starter bob29
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  • #1
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Homework Statement


Find the volume of the region in the 1st octant bounded above by the surface z=4-x^2-y and below by the plane z=3.
Answer = 4/15


Homework Equations


V = [tex]\int\int\int dV[/tex]


The Attempt at a Solution


I'm having trouble determining the upper and lower z limits.
I partly don't understand what to do, is the function z=3 or z=4-x^2-y that I am integrating.

[tex] \int_0^2 \int_0^{4-x^{2}} \int_0^3 \left (1) \right dz.dy.dx [/tex]
skipping the tedious integration working out.
V=16 which is not the right answer.

I just realised my diagram is wrong, as I drew it as a parabola concave down in the wrong plane.
I've written the new integral as
[tex] \int_0^2 \int_0^{4-x^{2}} \int_0^{3} \left (4-x^2-y) \right dz.dy.dx [/tex]
 
Last edited:

Answers and Replies

  • #2
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IMHO, this problem is easier with a double integral:

[tex]V = \iint_{\mathcal{D}} (z_2(x, y) - z_1(x, y))dA[/tex]​

where [tex]z_2(x, y) - z_1(x, y)[/tex] is the surface which is on top minus the surface which bounds the volume from below. That is, [tex]z_2(x, y) - z_1(x, y) = 4 - x^2 - y - 3 = 1 - x^2 - y[/tex]. The region of integration [tex]\mathcal{D}[/tex] is what the intersection of the two surfaces projects onto the first quadrant of the XY plane. That is, we set the z's of the two surfaces equal:

[tex]z_2(x, y) = z_1(x, y) \Rightarrow y = 1 - x^2[/tex]​

so the region bounded by [tex]x = 0[/tex], [tex]y = 0[/tex] and the graph of the function [tex]y = 1 - x^2[/tex].

[tex]\mathcal{D} \equiv \{(x, y) \in \mathbb{R}^2: x \geq 0,~ 0 \leq y \leq 1 - x^2 \}.[/tex]​

So, we have

[tex]V = \iint_{\mathcal{D}} (z_2(x, y) - z_1(x, y))dydx = \int_0^1 \int_0^{1 - x^2} (1 - x^2 - y) dydx[/tex]​

This integral is easy - I haven't done it, but I checked with my calculator and indeed, it does yield the answer you gave.

Good luck!

PS: If you notice, you can easily carry over to a triple integral, by adding an extra integral in [tex]z[/tex], with a lower limit of [tex]z = 3[/tex] and an upper limit of [tex]z = 4 - x^2 - y[/tex], and leaving rest of the double integral as it is. So,

[tex]V = \iiint_{\mathcal{V}} dzdydx = \int_0^1 \int_0^{1 - x^2} \int_3^{4 - x^2 - y} dzdydx[/tex]​

and you get the same double integral when performing the first integration in [tex]z[/tex].
 
Last edited:
  • #3
18
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Thanks for the help. I was wondering where I was going wrong, it was my x and y limits which I forgot to change. I'm not used to getting this type of question i.e where z=3.
 

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