Example: #10 (page 279)
| > | restart; with( plots ): with( Student[Calculus1] ): |
| > |
This problem was discussed in class on Monday, November 25. This discussion will be similar, except that Maple will be used to perform some of the computations leading to the formulation of the definite integrals. The problem is to find the area of the region bounded by the graphs of
,
, and the
-axis:
| > | eq1 := y = sqrt(x); eq2 := x + y = 6; eq3 := y = 0; |
| > | plot( [ rhs(eq1), solve(eq2,y), rhs(eq3) ], x=0..10, color=[red,blue,green], thickness=3 ); |
| > |
As discussed in class, this problem can be approached using either vertical or horizontal slices:
Vertical Slices
To use vertical slices it is necessary to identify the curves that comprise the top and bottom of the region as functions of x. The bottom of the region is always the
-axis:
| > | bottom := 0; |
The curve that defines the top of the region depends upon the value of
. The two top curves are
| > | top1 := solve( eq1, y ); |
| > | top2 := solve( eq2, y ); |
The transition between these curves occurs where they intersect:
| > | ip := solve( top1=top2, x ); |
| > |
The subregion where the square root function forms the top of the region corresponds to 0 <=
<= 4:
| > | area1 := Int( top1-bottom, x=0..4 ); |
The subregion where the linear function forms the top of the region corresponds to 4 <=
<= 6:
| > | area2 := Int( top2-bottom, x=4..6 ); |
The area of the full region is, therefore, the sum of these two subareas:
| > | area := area1 + area2; |
The value of these integrals is easily determined by Maple
| > | area = value( area ); |
Thus, this region has area 22/3.
| > |
Horizontal Slices
To use horizontal slices it is necessary to identify the curves that form the left and right sides of the region as functions of y. In this case, the left side is always formed by the graph of the square root function
| > | left := solve( eq1, x ); |
The right edge of the region is formed by the graph of the linear function:
| > | right := solve( eq2, x ); |
| > |
The typical (horizontal) slice has width
| > | width := left - right; |
and height Delta*y. These slices are formed for all value of y from zero up to the height where the two curves intersect:
| > | ip := solve( left=right, y ); |
The solution we need is the positive value of
. Thus, the area can be expressed as the definite integral
| > | area := Int( right-left, y=0..2 ); |
The value of this integral is
| > | area=value(area); |
| > |
| > |
Note that the value of the integral is the same,
, by either method.
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