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 August 31st, 2019, 10:11 AM #1 Newbie   Joined: Aug 2019 From: India Posts: 15 Thanks: 0 How to find the volume when solids intersect? I want to know how to find the volume when solids intersect. What I means is let the solids be :- $\displaystyle z= f(x,y)$ $\displaystyle y= g(x,z)$ $\displaystyle x= h(y,z)$ If I'm asked to find the volume contained by them then what is the thing that I shall have to integrate? I have found from my search that it is generally done by integrating $\displaystyle f(x,y) \times g(x,z) \times h(y,z) dx dy dz$. Here is the link from where I have inferred this https://math.stackexchange.com/quest...riple-integral Can anybody please explain how the integration of that product gives me the volume? Is my conclusion from that link wrong? The thing that I want to achieve is :to find the volume enclosed between solids without any visualization. Any help will be much appreciated.
 August 31st, 2019, 11:45 AM #2 Senior Member   Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 You can't define a volume as z=f(x,y). At best, you would get a surface. If you were looking at the solution with the Theta functions, what they were actually doing was creating functions $\displaystyle f_1(x,y,z), f_2(x,y,z), f_3(x,y,z)$ which equal one inside the solid, and zero outside. Multiplying them together gives you a function which is one inside the intersection of the three solids. Integrate this over $\displaystyle \mathbb{R}^3$, and you get the volume of the intersection. This is the type of approach we would use for numerically approximating the volume without worrying about the nature of the shapes. To get a closed-form solution, we typically have to solve for the boundaries of the intersection, and then just integrate dV between these limits. If the intersection is convex, then it's something like $\displaystyle \int_{z_{min}}^{z_{max}}\int_{y_{min}(z)}^{y_{max} (z)}\int_{x_{min}(y,z)}^{x_{max}(y,z)} dxdydz$. For concave shapes, we have to split up the integrals in places, or take advantage of symmetry, conics, or various other tricks. All of these are highly dependent on the shapes in play. Thanks from Adesh Mishra
August 31st, 2019, 07:19 PM   #3
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 Originally Posted by DarnItJimImAnEngineer You can't define a volume as z=f(x,y). At best, you would get a surface. If you were looking at the solution with the Theta functions, what they were actually doing was creating functions $\displaystyle f_1(x,y,z), f_2(x,y,z), f_3(x,y,z)$ which equal one inside the solid, and zero outside. Multiplying them together gives you a function which is one inside the intersection of the three solids. Integrate this over $\displaystyle \mathbb{R}^3$, and you get the volume of the intersection. This is the type of approach we would use for numerically approximating the volume without worrying about the nature of the shapes. To get a closed-form solution, we typically have to solve for the boundaries of the intersection, and then just integrate dV between these limits. If the intersection is convex, then it's something like $\displaystyle \int_{z_{min}}^{z_{max}}\int_{y_{min}(z)}^{y_{max} (z)}\int_{x_{min}(y,z)}^{x_{max}(y,z)} dxdydz$. For concave shapes, we have to split up the integrals in places, or take advantage of symmetry, conics, or various other tricks. All of these are highly dependent on the shapes in play.
When you said “multiplying together gives you a function which is one inside the intersection of the three solids” is the thing that I want to know. My question is : How and why the product gives us the function inside the instersection of solids? I request you kindly to please explain that part elaborately.

 September 1st, 2019, 05:53 AM #4 Senior Member   Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 The functions take on a value of 0 or 1, which is a mathematical way of representing a Boolean (true/false) value. Basically, f1 represents "inside shape 1?" True (1) or false (0) f2 is "inside shape 2?" True (1) or false (0) f3 is "inside shape 3?" True (1) or false (0) The ultimate question is "inside shape 1" AND "inside shape 2" AND "inside shape 3?" We get this by f1*f2*f3. If any one of them is false (0), we get 0. If all three are true (1), we get 1. This can also be extended to fuzzy logic (x AND y = x*y), by the way. Thanks from Adesh Mishra
 September 1st, 2019, 07:09 AM #5 Newbie   Joined: Aug 2019 From: India Posts: 15 Thanks: 0 Great explanation. Thank you.

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