When finding the volume of a solid using triple integrals, is the "integral" always int int int dV, regardless of what the "z function" is? For instance, for the problem: Find the volume of the solid that lies within the sphere x^2+y^2+z^2=4, above the xy-plane, and below the cone z=sqrt{x^2+y^2}. Is the integral just int int int dV with the correct bounds of integration, regardless of z=sqrt{x^2+y^2}?

Janiah Parks 2022-09-29 Answered
Volume and Triple Integrals
When finding the volume of a solid using triple integrals, is the "integral" always dV, regardless of what the "z function" is?
For instance, for the problem: Find the volume of the solid that lies within the sphere x 2 + y 2 + z 2 = 4, above the xy-plane, and below the cone z = x 2 + y 2 .
Is the integral just dV with the correct bounds of integration, regardless of z = x 2 + y 2 ?
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Answers (1)

falwsay
Answered 2022-09-30 Author has 8 answers
Explanation:
If you have triple integral of type F D V and region in some volume V. This means you are adding value of F over all points over V. But if F is 1, this means you're finding volume of V
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