# Thread: Math question - 3d integrals

1. Originally Posted by ubiquitous1
I think I understand what you are saying, but triple integrals are still used to find a volume.

V=∫∫∫ dz dy dx

Where the limits of integration for z are 0 to 1-y, limits for y are 0 to 1 and finally limits of x are 0 to 2.

I miss Calculus
That's how we used to do it, too, although it's conceptually rather primitive.

2. How you approach it depends on what you are solving. For A(h)=pi*h^2 (ie. a form of cone), you wouldn't need to perform a triple integral to find the volume. If you work it from basics it is a triple integral though, because the area came from integration over r and theta, then setting r=h.

So triple integrals are used to find volume from scratch, but are not always a necessary step if you recognise how the cross-sectional area behaves. That was the formula onemoretime had, which may or may not be useful depending on what needs to be solved.

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
Single Sign On provided by vBSSO