PLEASE HELP!! THIS IS DUE IN 13 HOURS!

[vvartani]

PLEASE HELP A COMP SCI MAJOR STUDENT OUT!!!!! (this is due in about 13 hours)
I need to Approximate the area of a quarter circle by using 10,000 rectangles.

I know I need to use riemann sum and I have tried adjusting this but I keep getting nan or some ridiculously high number. Ultimate test to see if this works is when I plug in a radius of 2 I should get pi. Please send help, it is desperately needed!
Thank you

#include <iostream>
#include <cmath>
using namespace std;
int main(){
int n=10000;
int r;
double x=r/n;
cout<<"Enter a radius for the circle: ";
cin>>r;
double area=0;
for(int i=1; i<=n; i+=1){
area += ((i*r)/n)*(sqrt((r*r)-((i*x)*(i*x))));
}
cout<<"The area of a circle with radius "<<r<<" is "<<area<<".";
return 0;
}

[lastchance]

r should be a double.
You can't work out x until AFTER you input r.
The widths of your rectangles are wrong - they should be constant (r/n) - and the abscissas are also biased to end of interval.

Use code tags.
Don't double post.

Should take you about 2 minutes to fix that, not 13 hours.

[kbw]

Solution seems wrong as I don’t see sin() or cos() anywhere in that sum; but of course, I am most probably wrong.

You seem to counting all rectangles,whereas you should be excluding those outside of the radius.

Checkout a Monte Carlo solution for PI, as it’s based on the same principle.

[lastchance]

He/she is (badly) numerically integrating the area under a quarter-circle x²+y²=r², or rearranged as
y=sqrt(r²-x²)

So, it should be a sum of rectangle areas: width (r/n) times height (sqrt(r²-x²)). (His/her x doesn't mean the same as mine. Under the square root, x=i*r/n is sort of OK, but is biased always to the end of the interval and in this instance where the height is decreasing will tend to give a slightly low answer.)

The area would then be (pi.r²/4), giving pi if r is 2.

Numerical integration is considerably more efficient than Monte Carlo.

[doug4]

Just a correction:

The area of a circle is pi*r^2. So if the radius is 2, the area would be 4*pi.

If you are given diameter, the area is (pi*d^2)/4. If the diameter is 2, the area would be pi.

[lastchance]

Just a correction:

The area of a circle is pi*r^2. So if the radius is 2, the area would be 4*pi.

Yeah, but this is a QUARTER circle.


As in (original post):

Approximate the area of a quarter circle by using 10,000 rectangles.

[doug4]

Boo on me. I obviously need remedial reading lessons.

Carry on.

[jonnin]

All I can say is...

https://www.pinterest.com/pin/567735096750286141/

[dhayden]

Break it up into variables so you don't get lost in the calculations.

#include <iostream>
#include <cmath>
using namespace std;
int
main()
{
    int n = 10000;
    double r;
    cout << "Enter a radius for the circle: ";
    cin >> r;
    double dx = r / n;   // the width of each rectangle
    double area = 0;
    for (int i = 1; i <= n; ++i) {
	double x = ???;	// Put in the formula for x at the right side of the rectangle
	double y = ???;      // Put in the formula for the the value of y at x.
	area += dx * y;      
    }
    cout << "The area of a circle with radius " << r << " is " << area << ".";
    return 0;
}

Here's what I get when I put in the correct(?) formulas:

Enter a radius for the circle: 2
The area of a circle with radius 2 is 3.14139.

[lastchance]

#include <iostream>
#include <cmath>

double Midpoint( double f( double ), double a, double b, int N )
{
   double sum = 0;
   double dx = ( b - a ) / N;
   for ( int i = 0; i < N; i++ ) sum += f( a + ( i + 0.5 ) * dx );
   return dx * sum;
}

int main()
{
   std::cout << Midpoint( []( double x ){ return sqrt( 4 - x * x ); }, 0, 2, 10000 ) << '\n';
}

[vvartani]

thank you everyone for your help, please forgive my obvious errors I am new to coding. I really appreciate all the help and I was definitely able to finish on time! Thank you all again.