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I am trying to understand volume and surface integrals. I do get the idea of the process (find a parametric equation of the volume/surface, integrate afterwards). But I just cannot make up parametric equations on my own.

As an example, one of my tasks is to find the volume of a space limited by the surfaces

$x=0, y=0, y=6, z=x^2, z=4$.

I pictured it and know what it looks like, but i just cannot think of any real basic approach to find the parametric equation. The 'general volume equation' we learned at uni looks like this:

$r(u,v,w)=x(u,v,w)e_1+y(u,v,w)e_2+z(u,v,w)e_3$ where $e_1,e_2,e_3$ are the unit vectors.

Does anybody have a hint or an idea to help me? Thank you in advance!

greets, Ari

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1 Answer 1

Here's one approach to the problem you gave:

When you envision the region, you should see that if you take a slice of the xz-plane anywhere within the domain, it will be identical. Therefore, it makes sense to take $y$ to be the outer integral, and then use the slice as your inner integral. (This is analogous to finding the volume of a cylinder by finding the area of its base(inner integral), and then multiplying by its height(outer integral)).

Now, back to your problem:

$$V = \int_0^6 \int_0^2 \int_{x^2}^4 dz dx dy$$ $$V = \int_0^6 \int_0^2 4-x^2 dxdy$$ $$V = \int_0^6 \big[4x-1/3x^3 \big]_0^2 dy$$ $$V = \int_0^6 16/3 dy$$ $$V = 32$$

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I found your answer illuminating!! Thanks a lot! –  ari Dec 8 '12 at 21:59

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