Integral Calculus?? Polar Curves?

Question

What is the area bounded by 2 polar curves??

r=2^(1/2)sinx

r^2=cos2x

with solution pls to check my answer

Answer 1

As I understand polar curves the radius "r" must be positive so in these equations x must be chosen so that this does not happen. The first is kind of an ellipse oriented along the y-axis in the first and second quadrants. The second curve is oriented along the x-axis and lies in the first and fourth quadrants. So the desired area is in the first quadrant and bounded on the top by r^2 = cos(2x) and on the bottom by r = 2^(1/2)sin(x).

r = 2^(1/2)sin(x)

r^2 = cos(2x)

Find the point of intersection:

[2^(1/2)sinx]^2 = cos(2x)

2sin^2(x) = cos(2x)

2{[1 - cos(2x)]/2] = cos(2x) ..... half angle formula

1 - cos(2x) = cos(2x)

cos(2x) = 1/2

2x = 60 degrees

x = 30 degrees

r = 2^(1/2)sin(30) ......... r = SQRT(2)/2

r^2 = cos(60) = 1/2 ...... r = SQRT(2)/2

So this looks good. Intersection point is r = SQRT(2)/2 and x = 30 deg

The area must be cut into two pieces. The bottom one will be the first equation integrated from 0 to pi/6. The top one will be the second equation integrated from pi/6 to pi/4.

Bottom section:

dA = r*(pi/6 - x)dr .... Integrated from r = 0 to r = SQRT(2)/2

x = arcsin[r/SQRT(2)]

dA = (pi/6)r*dr - r*arcsin[r/SQRT(2)]dr

A = (pi/12)r^2 - {r^2/2 - 1/2}arcsin[r/SQRT(2)] + rSQRT(2 - r^2)/4}

A = [pi/24 + (1/4)pi/6 - SQRT(2)SQRT(3/2)/8] - [0]

A = pi/12 - SQRT(3)/8 = 0.0453

Top section:

dA = r*(x - pi/6)dr .... Integrated from r = 0 to r = SQRT(2)/2

x = arccos(r^2)/2

u = r^2 then du = 2r*dr and r*dr = du/2

dA = (arccos(u)/2 - pi/6)(du/2)

dA = (1/4)[arccos(u) - pi/3]du

A = (1/4){[u*arccos(u) - SQRT(1 - u^2)] - pi*u/3}

Evaluate from r = 0 (u = 0) to r = SQRT(2)/2 (u = 1/2)

A = (1/4)[(1/2)(pi/3) - SQRT(3/4) - pi/6] - [-1/4]

A = -SQRT(3)/8 + 1/4 = 0.0335

Total area = pi/12 - SQRT(3)/8 - SQRT(3)/8 + 1/4

Total area = pi/12 - SQRT(3)/4 + 1/4

Total area =  0.0453 + 0.0335

Total area = 0.0788

Or do you mean the total area enclosed by both curves and not counting the overlap area twice. In this case the area will be the sum of the areas of both curves minus the overlap area (which was calculated above).

Area of r = 2^(1/2)sin(x)

dA = 2*r*(pi/2 - x)*dr  ..... integrate from r = 0 to r = SQRT(2)

dA = pi*r*dr - 2*r*arcsin[r/SQRT(2)] dr

A = pi*r^2/2 - 2*{r^2/2 - 1/2}arcsin[r/SQRT(2)] + rSQRT(2 - r^2)/4}

A = [pi/2] - [0]

A = pi/2

Area of r^2 = cos(2x)

dA = 2*r*x*dr .... Integrated from r = 0 to r = 1

x = arccos(r^2)/2

u = r^2 then du = 2r*dr and r*dr = du/2

dA = (arccos(u)/2)(du)

A = (1/2)[u*arccos(u) - SQRT(1 - u^2)]

When r = 0, u=0 and r = 1 and u = 1

A = [0] - [-1/2] = 1/2

Total area = pi/2 + 1/2 - overlap

Total area = pi/2 + 1/2 - 0.0788

Total area = 1.992