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