Double Integrals In Polar Coordinates Question?

Question

Find the area of the region contained by r = 1+sin(theta)

Answer 1

I haven't done one of these in quite a while.

What would you differential element be? How about r*(dtheta)*dr?

that sort of a ds*dr where ds depends on the radius. Try it!

Note that without the sin theta you'd get the int rdthetadr with r 0 to R and theta 0 to 2pi You'd get the correct area of a circle.

Try the intergral, see if it makes sense. Then sketch the figure and estimate it's area to compare. My confidence is increasing as I write this.

Answer 2

Hi Kath!
Let dS=dr*r/2 be the area of a narrrrrrow (elementary) triangle and dr=r*dt, or dS=(r*dt)*r^2/2=0.5*r^3*dt, then the whole area S=integral[for t=0 until 360] of 0.5*(1+sin(t))^3*dt – double integration is unnecessary!
(1+sint)^3=1+3sint+3(sint)^2+(sint)^3
S=S1+S2+S3, S1=t-3cost=2*pi,
S2=I[for t=0 until 360] of 3(sint)^2*dt,
S3=I[for t=0 until 360] of (sint)^3*dt.
Mind! (sint)^2=0.5*(1-cos2t), thus S2=3*pi.
Mind! (sint)^3=sint*(1-(cost)^2), thus S3=0. The total S=5*pi