# Find the area of a loop of the curve r=a sin3theta?

#### Risposta:

(pia^2)/12

#### Spiegazione:

Dove a=1, the curve looks like: Increasing or decreasing the value of a will only change the radius of the curve.

To find when the curve begins and ends, set r=0, since this is where the curve is at the origin.

If asin3theta=0, poi sin3theta=0. Da sintheta=0 at theta=0,pi,2pi... we see that for sin3theta, sarà 0 at 0,pi//3,2pi//3...

So, the curve in the first quadrant varies from theta=0 a theta=pi//3.

The expression for the area of any polar equation r da theta=alpha a theta=beta è dato da 1/2int_alpha^betar^2d theta.

For one loop of the given equation, the corresponding integral is then 1/2int_0^(pi//3)(asin3theta)^2d theta.

Working this integral:

1/2int_0^(pi//3)(asin3theta)^2d theta=1/2int_0^(pi//3)a^2(sin^2 3theta)d theta

Usa l'identità cos2alpha=1-2sin^2alpha to rewrite for sin^2alpha, Mostrando che sin^2alpha=1/2(1-cos2alpha).

We can use this to say that sin^2 3theta=1/2(1-cos6theta). Then the integral reduces:

=1/2int_0^(pi//3)a^2(1/2(1-cos6theta))d theta=a^2/4int_0^(pi//3)(1-cos6theta)d theta

Integrating term by term:

=a^2/4(theta-1/6sin6theta)|_0^(pi//3)

=a^2/4[pi/3-1/6sin2pi-(0-1/6sin0)]

=a^2/4(pi/3)

=(pia^2)/12