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Doesn't the question give 3 equations as constraints for the volume enclosed? Can you do a 3-D graph of all 3 equations?stolencookie said:
Can't do a 3D graph the two constraints are y=x and y=3x-x^2 , I use the z=x^2+xy to find the volume using the double integrals just having trouble with the set up.berkeman said:
I would assume you want the upper region. The lower region has a boundary portion of ##y=0## which is not mentioned in the problem.stolencookie said:
stolencookie said:
The volume of a double integral allows us to calculate the volume of a three-dimensional region that is bounded by two curves or surfaces. This is useful in many scientific fields, such as physics, engineering, and economics.
To find the region using graphed equations, we first need to plot the equations on a graph. The intersection points of the curves or surfaces will determine the boundaries of the region. Then, we can set up the double integral using these boundaries to calculate the volume of the region.
A single integral is used to find the area under a curve, while a double integral is used to find the volume between two curves or surfaces. A double integral involves integrating with respect to two variables, whereas a single integral only involves one variable.
Yes, the volume of a double integral can be used to find the volume of irregular shapes as long as the equations that define the boundaries of the region can be graphed and the intersection points can be determined.
The volume of a double integral has many applications in physics, engineering, and economics. For example, it can be used to calculate the volume of a solid object, the displacement of a fluid, or the value of a complex financial asset. It is also used in the calculation of moments of inertia and center of mass in physics.
