## anonymous 2 years ago Hi all! Need serious Multivariable Calc help. Right now we're learning practical applications of double integrals: An annulus with inner radius r=1 and outer radius r=2 has density equal to delta(x,y)=3*(y+sqrt(x^2+y^2))/pi. Compute the mass of the annulus using polar coordinates. I get an answer, but I'm almost sure it's wrong.

1. anonymous

to find the area we must integrate $$\delta$$ over the annulus, which is clearly represented in polar coordinates using $$\{(r,\theta):1\le r\le 2\}$$

2. anonymous

so notice we must rewrite $$\delta$$ in terms of $$r,\theta$$; this is no problem as $$r=\sqrt{x^2+y^2}$$ and $$y=r\cos\theta$$:$$\delta(r,\theta)=\frac3\pi r\ (1+\cos\theta)$$

3. anonymous

now we set up our integral:$$m=\int_0^{2\pi}\int_1^2\frac3\pi r\ (1+\cos\theta)\ dr\ d\theta=\frac3\pi\left(\int_0^{2\pi}(1+\cos\theta)\ d\theta\right)\left(\int_1^2r\ dr\right)$$

4. anonymous

so clearly $$\displaystyle\int_1^2r\ dr=\frac12(2^2-1^2)=\frac32$$ and the other integral turns out to be rather simple as well: $$\int_0^{2\pi}(1+\cos\theta)\ d\theta=\bigg[\theta+\sin\theta\bigg]_0^{2\pi}=2\pi+\sin2\pi-\sin0=2\pi$$

5. anonymous

thus $$\displaystyle m=\frac3\pi\cdot\frac32\cdot2\pi=9$$

6. anonymous

Oh wow Thank you so much!!!

7. anonymous

notice it's safe to rewrite the integral as I did above:$$\int\int f(x)g(y)\ dx\ dy=\int g(y)\int f(x)\ dx\ dy=\int f(x)\ dx\cdot\int g(y)\ dy$$

8. anonymous

yeah we saw that in class about 2 days ago.

9. anonymous

only because there were no factors that depended on both $$r,\theta$$... the $$1+\cos\theta$$ only depends on $$\theta$$ so we may pull it out as a constant from the $$dr$$ integral; similarly, $$r$$ depends not on $$\theta$$ and so we may pull it out of the $$d\theta$$ integral

10. anonymous

ok, just wanted to clarify... I do it all the time and I know it confuses the people I tutor IRL

11. anonymous

@oldrin.bataku wait...isn't y = r sin theta?

12. anonymous

oops -- good catch. I don't think it'll make a difference though

13. anonymous

ok I'll double check the work and get back 2 ya

14. anonymous

it won't make a difference since $$\cos(t)$$ and $$\sin(t)$$ have the same area over $$[0,2\pi]$$... all this means is I picked a slightly different coordinate system:|dw:1389663840845:dw|this is the regular polar coordinate system

15. anonymous

here's my coordinate system:|dw:1389663889927:dw|

16. anonymous

oh that makes sense. just another question: when converting to a polar integral, doesn't dxdy become rdrdtheta?

17. anonymous

so the final answer is 14.

18. anonymous

oops, correcting for that we get $$\int_1^2 r^2\ dr=\frac13(2^3-1^3)=\frac73$$ergo our final result is $$\displaystyle\frac3\pi\cdot\frac73\cdot2\pi=14$$

19. anonymous

sweet. Thanks.