Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant between the circles x^2+y^2=64 and x^2 - 8x + y^2 = 0.
14 pi
the answer u gave me is incorrect. and please tell me the method u tried
To evaluate the given integral using polar coordinates, we first need to convert the given equations of the circles into polar form.
The equation of the circle x^2 + y^2 = 64 can be written in polar coordinates as r^2 = 64.
Next, we rewrite the equation x^2 - 8x + y^2 = 0 as:
(x^2 - 8x) + y^2 = 0
Completing the square, we have:
(x^2 - 8x + 16) + y^2 = 16
(x - 4)^2 + y^2 = 16
This equation can be written in polar form as (r*cos(theta) - 4)^2 + (r*sin(theta))^2 = 16.
Now, we need to find the limits of integration in terms of theta.
In the first quadrant, theta ranges from 0 to π/2.
To find the limits of integration for 'r', we need to determine the distance between the two circles at a given theta.
Setting the two equations for r equal to each other, we have:
r^2 = 64
(r*cos(theta) - 4)^2 + (r*sin(theta))^2 = 16
Simplifying the second equation:
r^2*cos^2(theta) - 8r*cos(theta) + 16 + r^2*sin^2(theta) = 16
r^2(cos^2(theta) + sin^2(theta)) - 8r*cos(theta) = 0
r^2 - 8r*cos(theta) = 0
r(r - 8*cos(theta)) = 0
From this equation, we have two possible values for r:
r = 0 (the origin)
r = 8*cos(theta)
Since we want to find the area between the two circles, the limits of integration for 'r' will be from 0 to 8*cos(theta).
The integral that gives the area can be written as:
A = ∫[0,π/2] ∫[0,8*cos(theta)] r dr d(theta)
Evaluating this double integral will give you the area between the two circles in the first quadrant.
You don't need Calculus to do this.
The outer circle has centre at (0,0) and has radius 8
The other circle has centre (4,0) and radius 4, so the touch at (8,0)
The big circle has area 64π, so the area in the first quad is 16π
The smaller circle has area 16π and half of it lies in quad I, so the area in there is 8π
So the area between the big circle and the little circle in quad I = 16π-8π = 8π