Polar coordinates, interval of integration?

Question

1) Why is the interval of integration for dθ from just 0 to π?

It interval kinda looks like it's from -π/4 to π/4. So I can see the "duration" of the integral is π radians also. So...

2) Why does it not matter where we start our integration? What am I missing?

Ẩn danh · Accepted Answer

(r*cos(θ) - 1)^2 + ((r*sin(θ))^2) = 1;

(r*cos(θ) - 2)^2 + ((r*sin(θ))^2)= 4;

r = 2*cos(θ);

r = 4*cos(θ);

Limits are from -π/2 <= θ <= π/2;

int_(-π/2)^(π/2) int_(2*cos(θ))^(4*cos(θ)) int_0^(4 - ((r*cos(θ) - 2)^2) - ((r*sin(θ))^2)) r dz dr dθ = 11π/2

It just so happens when you do the limits of integration for this case it works out:
m = θ + (π/2); dm = dθ; m - (π/2) = θ

int_(0)^(π) int_(2*cos(m - (π/2)))^(4*cos(m - (π/2))) int_0^(4 - ((r*cos(m - (π/2)) - 2)^2) - ((r*sin(m - (π/2)))^2)) r dz dr dm;

Let m = θ; dm = dθ;

int_(0)^(π) int_(2*sin(θ))^(4*sin(θ)) int_0^(4 - ((r*sin(θ) - 2)^2) - ((-r*cos(θ))^2)) r dz dr dθ =

int_(0)^(π) int_(2*sin(θ))^(4*sin(θ)) int_0^(4 - ((r*sin(θ) - 2)^2) - ((r*cos(θ))^2)) r dz dr dθ = 11π/2

? · Answer

The interval -π/2 to ��/2 is also a valid description of the circles. The positive r values over (-π/2,0) describe the same (x,y) points as the negative r values over (π/2, π).

Without solving the entire problem, I'll  guess that using the 0 bound makes the integral easier.

For a test, you just need to memorize the equations of some simple graphs such as a circle.

Polar coordinates, interval of integration?

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