How do you evaluate the integral sin(x2+y2)dr where r is the region 9x2+y264 in polar form?

1 Answer
Feb 26, 2015

We should indeed use the polar coordinate system:
x=rcosϕ
y=rsinϕ

When you substitute these in x2+y2, you get
x2+y2=(rcosϕ)2+(rsinϕ)2
=r2cos2ϕ+r2sin2ϕ
=r2(cos2ϕ+sin2ϕ)
=r2
for cos2+sin2=1

Our area becomes 9r264 by substituting x2+y2=r2.
Therefore, our limits for r are: 3r8, by taking the square root.

When we use polar coordinates, we should take the Jacobian(=J(r,ϕ)) into account, which is equal to r for polar coordinates. Our integral becomes:
sin(x2+y2)dr=sin(r2)J(r,ϕ)dr=83rsin(r2)dr

To solve this integral, we need to use substitution. We substitute u=r2, so du=2rdr, in other words dr=12rdu.
83rsinr2dr=649rsinu12rdu=64912sinudu=[12cos(u)]649=cos(9)cos(64)2