BuyFind

Essential Calculus: Early Transcen...

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

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

Answer to Problem 1RCC

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

Answer to Problem 1RCC

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

To write: The geometric interpretation of Rf(x,y)dA when f0.

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

Answer to Problem 1RCC

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

To interpret: About the Midpoint Rule for double integrals.

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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