Multivariable CalculusMultivariable integration
Integrating a function is a way of totaling up its values. For example, if is a function from a region in to which represents the mass density of a solid occupying the region , we can find the total mass of the solid as follows: (i) split the region into many tiny pieces, (ii) multiply the volume of each piece by the value of the function at some point on that piece, (iii) and add up the results. If we take the number of pieces to infinity and the piece size to zero, then this sum converges to the total mass of the solid.
We may apply this procedure to any function defined on , and we call the result the integral of over , denoted .
To find the integral of a function defined on a 2D region , we set up a double iterated integral over : the bounds for the outer integral are the smallest and largest possible values of for point in , and the bounds for the inner integral are the smallest and largest values of for any point in a given each " = constant" slice of the region (assuming that each slice intersects the region in a line segment).
Find the integral over the triangle with vertices , , and of the function .
Solution. The least and greatest values of for any point in the region are 0 and 3, while the least and greatest values of for each given -slice are 0 and . Therefore, the integral is
To set up an integral of a function over a 3D region (for the order ): the bounds for the outer integral are the smallest and largest values of for any point in the region of integration, then the bounds for the middle integral are the smallest and largest values of for any point in the region in each " = constant" plane, and the inner bounds are the smallest and largest values of for any point in the region in each " = constant" line.
Integrate the function over the tetrahedron with vertices , , , and .
Solution. The least and greatest values of are 0 and 4, so those are our outer limits (see the figure below). For a fixed value of , the least and greatest values of for a point in are 0 and , respectively. Finally, for fixed and , the least and greatest values of for a point in are 0 and the point on the plane with the given values of and . So we get
Evaluate by writing it as an integral over a region in the plane and then integrating over the region with respect to the opposite order of integration.
Solution Let's begin by drawing the region:
We can see that this is the region under the graph of from to . Thus we integrate as ranges from 0 to 2 and (for each fixed value of ) as ranges from 0 to . We get
Consider the region between the parabolas and . Find .
Solution. We see that the curves intersect at and . So we get