Integration of parabola

Question

I have this homework question I am working on:

The base of a sand pile covers the region in the xy-plane that is bounded by the parabola $x^2 +y = 6$ and the line $y = x$: The height of the sand above the point $(x;y)$ is $x^2$: Express the volume of sand as (i) a double integral, (ii) a triple integral. Then (iii) find the volume.

I have drawn the $x^2 + y = 6$ and $y=x$ plane and found the intersection between the functions to be $(-3,-3)$ and $(2,2)$. So I now know what the base looks like. Now I am REALLY confused what the question means about the $x^2$ being the height. What point are they talking about?

Also, if it is a volume then doesn't it HAVE to be a triple integral? How can I possibly express it as a double integral?

A triple integral over a set, of the constant function 1, gives the volume of the set. A double integral of a (positive) function of 2 variables gives the volume under the graph of that function. — mathematician, Nov 29 '13 at 10:00
@mathematician ahhh I see what you mean! This perspective will take a little more thinking on my end to sink in. — Raynos, Nov 29 '13 at 12:01

Mikasa · Accepted Answer · 2013-11-29T10:50:00.643

2

$$\int_{-3}^{2}\int_{y=x}^{y={6-x^2}}z dydx,~~~z=x^2$$ Or $$\int_{-3}^{2}\int_{y=x}^{y={6-x^2}}\int_{z=0}^{x^2}f(x,y,z) dzdydx,~~~f(x,y,z)=1$$

enter image description here

edited Nov 29 '13 at 10:50

answered Nov 29 '13 at 10:18

Mikasa

67,374

may I ask what software you use to graph this? I think it would really help me in integral problems. – Raynos Nov 29 '13 at 10:57
@Raynos: I used Maple 16 for these kinds of problems. :-) – Mikasa Nov 29 '13 at 11:14
Ahhh! I disliked maple because I am not familiar with the syntax...I didn't know it was so cool. Do you know how I can start learning the maple syntax? – Raynos Nov 29 '13 at 11:21
@Raynos: I think there are some online handbooks about it. I think you'll change your mind after knowing it. :-) – Mikasa Nov 29 '13 at 11:35
Very pretty graphic, too! +1 – amWhy Nov 29 '13 at 13:42

Cure · Answer 2 · 2013-11-29T10:13:31.913

Now I am REALLY confused what the question means about the x2 being the height. What point are they talking about?

Means exactly that, the height of the surface on the $z$ axis is given by $z=x^2$, you can also look at it like a function on the $xy$ plane given by $z=f(x,y) = x^2$.

Also, if it is a volume then doesn't it HAVE to be a triple integral? How can I possibly express it as a double integral?

No. Take single integrals as example, you can determine the lenght of a path (1 dimension), areas (2 dimensions) or even some volumes (solids of revolution = 3 dimensions).

Generally, you can find the volume of a closed region with $\int\int\int dV$ or an integral of the form $\int\int f(x,y)\;dS$ -asuming that the region is bounded by the surface given by $z=f(x,y)$-.

Thank you for the great explanation. – Raynos Nov 29 '13 at 11:21 — Raynos, Nov 29 '13 at 11:21

Integration of parabola

2 Answers2