Finding Volume Using Triple Integrals with Non-Standard Limits

[bob29]

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 = $$\int\int\int dV$$

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.

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

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

 

[Metaleer]

IMHO, this problem is easier with a double integral:

$$V = \iint_{\mathcal{D}} (z_2(x, y) - z_1(x, y))dA$$​

where $$z_2(x, y) - z_1(x, y)$$ is the surface which is on top minus the surface which bounds the volume from below. That is, $$z_2(x, y) - z_1(x, y) = 4 - x^2 - y - 3 = 1 - x^2 - y$$. The region of integration $$\mathcal{D}$$ 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:

$$z_2(x, y) = z_1(x, y) \Rightarrow y = 1 - x^2$$​

so the region bounded by $$x = 0$$, $$y = 0$$ and the graph of the function $$y = 1 - x^2$$.

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

So, we have

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

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 $$z$$, with a lower limit of $$z = 3$$ and an upper limit of $$z = 4 - x^2 - y$$, and leaving rest of the double integral as it is. So,

$$V = \iiint_{\mathcal{V}} dzdydx = \int_0^1 \int_0^{1 - x^2} \int_3^{4 - x^2 - y} dzdydx$$​

and you get the same double integral when performing the first integration in $$z$$.

 

[bob29]

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.