Using polar coordinates, evaluate the integral ∫∫rsin(x2+y2)da where r is the region 16≤x2+y2≤81.

In polar coordinates, [tex]x^2+y^2=r^2[/tex], so the integral can be written as

[tex]\displaystyle\iint_R\sin(x^2+y^2)\,\mathrm dA=\int_{\theta=0}^{\theta=2\pi}\int_{r=4}^{r=9}r\sin(r^2)\,\mathrm dr\,\mathrm d\theta[/tex]
[tex]=\displaystyle\pi\int_{s=16}^{s=81}\sin s\,\mathrm ds[/tex]

where we take [tex]s=r^2[/tex] so that [tex]\mathrm ds=2r\,\mathrm dr[/tex].

[tex]=\pi(\cos16-\cos81)[/tex]

Answer Link

sheenuvt12 sheenuvt12 · Answer 2 · 2022-06-01T23:57:35+02:00

The value of the integral is: πcos(16)- πcos (81)

What are polar coordinates?

The polar coordinate system is a two-dimensional coordinate system in which each point on a plane is determined by a from a reference point and an angle from a reference direction.

Given:

∫∫ r sin(x²+y²)dA and 16≤x²+y²≤81.

Now,

x =rcosϕ

y=rsinϕ

I=r

Find the integration limits:

ϕ∈[0,2π]

16≤x²+y²≤81

16≤r² cos² Ф + r²sin² Ф≤81

16≤r²≤81

4≤r≤9

So, r∈[4,9]

Now,

∫∫ r sin(x²+y²)dA=

= [tex]\int\limits^9_4 {dr} \int\limits^{2\pi}_0 {rsin \;(r^{2}) } \, d\phi[/tex]

=2π [tex]\int\limits^9_4 {rsin \;(r^{2}) } \, dr[/tex]

Let r² = u

2rdr=du

Also, limit will change from 16 to 25.

5→25

4→16

So,

= π[tex]\int\limits^{81}_{16} {sin \;(u) } \, du[/tex]

=−π|cos (u)∣[tex]^{81}_{16}[/tex]

= πcos(16)- πcos (81)

Learn more about integral here:

https://brainly.com/question/17191581

#SPJ2