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Applications of integral calculus?
Integration can be used to calculate the area under a curve and the volume of solids of revolution.
What is the importance of integral calculus in engineering?
Its importance is tremendous - it has many different applications. Some of the applications include calculation of area, of volume, moment of inertia, of work, and many more.
What is a term used in math'?
volume integral
A derivative is to the rate of change as an integral is to?
A derivative is to the rate of change asan integral is to area/volume.
Is integration a way of calculating the volume of the same function equation with an additional dimension or is it always completely new volume rotated about an axis ie a circle to a sphere?
Geometrically the definite integral from a to b is the area under the curve and the double integral is the volume under the surface. So just taking the integral of a function does not yield the volume of the solid made by rotating it around an axis. An integral is only a solid of revolution if you take an infinite sum of infinitesimally small cylinders that is the disk method or you do the same with shells.
Applications of integral calculus?
Integration can be used to calculate the area under a curve and the volume of solids of revolution.
What is the importance of integral calculus in engineering?
Its importance is tremendous - it has many different applications. Some of the applications include calculation of area, of volume, moment of inertia, of work, and many more.
What is a term used in math'?
volume integral
A derivative is to the rate of change as an integral is to?
A derivative is to the rate of change asan integral is to area/volume.
What is the difference between triple integral and volume integral?
While I searching for the answer to this question, I totally confused. Atlast I reach in one thing that we may compute some volume integrals by using double integral but to evaluate a triple integral we should go through all the three integrals.
Why is biotechnology so important?
Because its applications are part of the integral elements in our society..
Is integration a way of calculating the volume of the same function equation with an additional dimension or is it always completely new volume rotated about an axis ie a circle to a sphere?
Geometrically the definite integral from a to b is the area under the curve and the double integral is the volume under the surface. So just taking the integral of a function does not yield the volume of the solid made by rotating it around an axis. An integral is only a solid of revolution if you take an infinite sum of infinitesimally small cylinders that is the disk method or you do the same with shells.
Application of definite Integral in the real life Give some?
What are the Applications of definite integrals in the real life?
What can you find with triple integrals area or volume?
A triple integral will usually give a measure of volume in 4-dimensional hyper-space.
Discuss line integral and its real life applications?
A line integral can evaluate scalar and vector field functions along a curve/path. When applied on vector field, line integral is considered as measure of the total effect of the vector field along a specific curve whereas in scalar field application, the line integral is interpreted as the area under the field carved out by a particular curve.Line integral has many applications in physics. In mechanics, line integral is used to determine work done by a force in moving an object along a curve. In circuit analysis, it is used for calculating voltage.
What is the method to determine the volume of a very small material?
A triple integral should suffice to locate the volume. Provided you can determine a equation to bound the surface.
You want derivation of cone volume?
This can be done easiest with integrals. Divide the cone into cylinders of very small height, and add up the volume of all the cylinders. Writing that as an integral will immediately give you the volume of the cone. For example, and for the simplifying case that the height is equal to the radius (r), you get the integral from x = 0 to r, of pi*x2. This integral is [(1/3)x3] with an upper limit of r and a lower limit of 0, or simply (1/3)r3.
