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Finding the volumes of a rectangular solid, prism, and pyramid Find the volume of (a) A rectangular solid having a length of 6 in, a width of 4 in, and a height of 1 ft (b) A prism having a height of 15 yd and a triangular base of 120 ft2 (c) A pyramid having a height of 8 cm and a square base whose side is 41 cm 2

( m=--

6(4)(12) 120(45)

Radius = Sqr(Area / 3.141593)

We use doubl e precision floats to minimize roundoff error: main0 { int n; // number of data points double x, y, sumX = 0.0, sumY = 0.0, sumXX = 0.0, sumXY = 0 0; . tout << "How many points: '; tin >> n; tout << "Enter ' << n C-C ' pairs, one pair per line:\n"; for (int i = 1; i <= n; i++) tout << I\, c-c i << 'I: '1; tin >> x >> y; sumX += x; // accumulate the sum of the x's in sumX sumY += y; // accumulate the sum of the y's in sumY sumXX += x*x; // accumulate the sum of x*x in sumXX sumXY += x*y; // accumulate the sum of x*y in sumXY double meanX = sumX/n; double meanY = sumY/n; double m = (sumXY - meanY*sumX)/(sumXX meanX*sumX); double b = meanY - meanX*m; tout << "The equation of the regression line is:\n' "\ty = ' << m C-K 'lx + ' C-C b << end10I

1 9 2 3 2 (8)

Each of the four sums sumX, sumY, sumXX, and sumXY is accumulated within the input loop. Then the averages meanX and meanY are computed. Then they are used in the formula to-compute the slope m and the y-intercept b of the regression line. The output from this program is very useful. The regression line is the straight line that best fits the given data. That is, among all possible straight lines, the one given by the equation y=1111.01x + 4444.03

is the best fitting line, in the sense that the sum of the squ.ares of the y-distances from the data points to the line is minimal. The value of this result is that it can be used for interpolation (and extrapolation). For example, to guess at the probable y-value corresponding to the x-value 3.2, simply substitute that into the following equation: y = 111 l.Ol(3.2) + 4444.03 = 3555.03 + 4444.03 = 7999.26.

Finding the volumes of a sphere, cylinder, and cone Find the volume of (a) A sphere with a radius of 10 in (b) A cylinder with a height of 4 yd and a base whose radius is 2 ft (c) A cone with a height of 2 ft and a base whose radius is 2 yd

The Monte Carlo simulation method is named after the casino in Monaco. It consists of picking points at random and counting those that satisfy certain criteria. It can be used to compute n: by simulating the tossing of darts at a circular dart board mounted on a square:

4 3 3 pr 4 3 3 (3.14)10 2 3

If the darts are equally likely to hit any point in the square, then the proportion that hit inside the circle will approximate the ratio of the area of the circle to that of the square. If the square has sides of length 2.0, then that ratio is (n: )/(s2) = (n: 1.02)/(2.02) = 7c/4, so 4 times that ratio will approximate K. It is easier to use the quarter circle of radius 1.0 that lies in the first quadrant. This way, the ran-

(3.14)(22)12

domly selected coordinates will all be in the range 0.0 to 1 .O. The area of the square is 1 .02 = 1 .O and the area of the quarter circle is (II: 1 .02)/4 = n/4, so the ratio is still 7c/4. #include #include #include main0 const long int tosses = 1000; // toss 1000 darts long int hits = 0; float x, y; unsigned seed = time(NULL); srand(seed); for(long int i = 0; i < tosses; i++) x = float(rand())/RAND -MAX; Y = float(rand())/RAND -MAX; if (x*x + y*y -C 1) ++hits; tout -CC 4.0*hits/tosses -CC endl; <iostream.h> <stdlib.h> ctime.h>

1 2 3 (3.14)(6 )(2)

Notice that the argument of Sqr is the numeric expression (Area / 3.141593). Of course, we could also have written

3.. $.35.04