Figure 1: Question 2 show code, possible partial credit . Instead of a unit circ

Question

Figure 1: Question 2 show code, possible partial credit . Instead of a unit circle described by the equation r2+92 # 1. it is al o possible to have curves 2 described by the equation Here a, ? > 0 and ? * ß in general-The curve with (:: 1.7 and ? 3.2 is plotted in Fig.1. You are roquired to numerically calculate the area enclosed by the curve r7+21 1. Let's not panic and be stupid and think this is a tro-dimensional integral 2. Observe that the curve is symmetric around both the z and y axes 3. Hence we calculate the area in the first quadrant r,g2 0 and multiply the result by 4 . The Cartesian plane is divided into four quadrants 5. See Fig, 1. (a) First quadrant: top right?26. 320. (b) Second quadrant: top left 0,y20 (c) Third quadrant: botton left r so,s d) Fourth quadrant: bottom right0,ys The area A under the curve in the finst quadrant is given boy the following integral: (2.2) Write down the equation for p(a) in the first quadrant. Prove that the integral in eq. (2.2) is a proper integral (a) midpoint rule, (b) trapezoid rule, (c) Simpson's rule. .Compute the value of Ai using Use -1000 subintervals for all three calculations 2. You are permitted to use the code displayed in the online lecture notes. ·The total area A is given by A = 4A1. ·State your computed values for the total area A to for decimal places. Denote your values by A, A As . Do not worry if the three resalts you obtain are not egual to four decimal places

Explanation / Answer

PROGRAM CODE:

#include<conio.h>

#include<graphics.h>

int firstquad(int a,int b,int c);

int midpoint(int x ,int y, int ,r);

void main()

{

clrscr();

int gdriver=DETECT,gmode;

initgraph(&gdriver,&gmode,"C:TCBGI");

int x,y,r;

cout<<"enter circle coordinates";

cin>>x>>y>>r;

circle(x,y,r);

firstquad(x,y,r);

midpoint(x,y,r);

getch();

closegraph();

}

int firstquad(int x,int y,int r)

{

int n;

n=x+r; //coefficients of x and y are 1 as it is a unit circle and thus the equation

int y1=n;

if(y1!=NAN)

cout<<" the given integral is a proper integral";

else

cout<<" it is not a proper integral";

return y1;

}

int midpoint(int x,int y,int r)

{

int d=3-(2*r);

if(d>0)

{

x=x+1;

y=y-1;

cout<<x<<y<<"is the mid point";

}

else{

x=x+1;

y=y;

cout<<x<<y<<"is the mid point";

}

return 0;

}

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