# An expression for a double Riemann sum of the given function. ### Essential Calculus: Early Transcen...

2nd Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781133112280 ### Essential Calculus: Early Transcen...

2nd Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781133112280

#### Solutions

Chapter 12, Problem 1RCC

(a)

To determine

## To write: An expression for a double Riemann sum of the given function.

Expert Solution

The expression for a double Riemann sum is i=1mj=1nf(xi*,yj*)ΔA_.

### Explanation of Solution

Given that the continuous function f is defined on a rectangle R=[a,b]×[c,d].

The double integral of f over the rectangle R is given by,

Rf(x,y)dAlimm,ni=1mj=1nf(xi*,yj*)ΔA

Here, ΔA=lb, where l,b are the length and breadth of each rectangle.

The given continuous function is f(xi*,yj*).

The sample points of each rectangle is denoted by (xi*,yj*).

The image value of the sample points under the function f is denoted by f(xi*,yj*) and the Riemann sum constants are denoted by m, n.

The sum mentioned above i=1mj=1nf(xi*,yj*)ΔA_ is the double Riemann sum of f. If f0 and m,n tends to infinity, then the double Riemann sum approximates to the volume under the given surface.

(b)

To determine

### To write: The definition of ∬Rf(x,y)dA as a limit.

Expert Solution

The definition of Rf(x,y)dA can be expressed as limm,ni=1mj=1nf(xi*,yj*)ΔA_.

### Explanation of Solution

The double integral can be expressed in terms of double Riemann sum as follows:

The double integral of f over the rectangle R is,

Rf(x,y)dAlimm,ni=1mj=1nf(xi*,yj*)ΔA

Here, ΔA=lb, where l,b are the length and breadth of each rectangle.

The given continuous function is f(xi*,yj*).

The sample points of each rectangle is denoted by (xi*,yj*).

The image value of the sample points under the function f is denoted by f(xi*,yj*) and the Riemann sum constants are denoted by m, n.

Thus, the definition of Rf(x,y)dA can be expressed as limm,ni=1mj=1nf(xi*,yj*)ΔA_.

(c)

To determine

Expert Solution

### Explanation of Solution

When f0, the double integral denotes the volume of the surface above the xy-plane and below the given function. The formula for finding this is given above in part (b).

If suppose the given function f takes both positive and negative values, then it does not denote the volume exactly. But, it is taken that the volume of the function of the two graphs one above the xy-plane and one below the xy-plane.

(d)

To determine

### To evaluate: The value of the double integral ∬Rf(x,y)dA.

Expert Solution

The value of Rf(x,y)dA is Rf(x,y)dA=abcdf(x,y)dydx_.

### Explanation of Solution

Rewrite the indefinite double integral by definite double integral from the equations or inequalities in the given rectangle. Then, as per the rules of integration, integrate it to get the value of the given double integral. That is,

Rf(x,y)dA=abcdf(x,y)dydx.

Thus, the value of Rf(x,y)dA is Rf(x,y)dA=abcdf(x,y)dydx_.

(e)

To determine

Expert Solution

### Explanation of Solution

The double integral, Rf(x,y)dAi=1mj=1nf(x¯i,y¯j)ΔA,

Here, ΔA=lb , where l, b are the length and breadth of each rectangle.

The given function is f(x,y).

The mid points of each rectangle is denoted by (x¯i,y¯j), where x¯i is the midpoint of [xi1,xi] and y¯i is the midpoint of [yj1,yj].

The Riemann sum constants are denoted by m, n.

Separate the given region by small rectangles by the method of Riemann sum for the double integrals. Then, pick the sample points from the Midpoint of each rectangle.

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