## anonymous 4 years ago Evaluate the integral ∬sin(x^2+y^2 )dA where R is the region that lies above the x-axis within the circle x^2 +y^2 =16 by changing into polar coordinates.

1. anonymous

as for radius is from -4 to 4. for angle is from 0 to 180 ?

2. experimentX

yep!!

3. anonymous

differentiate by r dr dθ ?

4. experimentX

use radiant (0 to pi)!! and dA = r dr d(theta)

5. anonymous

ok. let me work out....

6. experimentX

$\int_0^4 \int_0^\pi \sin ( r^2 \cos^2 \theta + r^2 \sin^2 \theta) \; r \; dr\; d\theta$

7. anonymous

could u pls explain to me why e radius is from 0 to 4 nt -4 to 4? and hw to get e function?

8. anonymous

ok i understood. hw abt e integration part?

9. experimentX

$\int_0^4 \int_0^\pi \sin ( r^2) \; r \; d\theta\; dr\ \\ \int_0^4 \sin ( r^2) \; r \; \left [ \theta \right ]_0^\pi dr\\$ just use subs r^2 = u

10. anonymous

ok...thanks. let me work out

11. anonymous

is the ans $- \pi \cos8 12. anonymous \[-\pi \cos 8$

13. anonymous

am i right??

14. experimentX

http://www.wolframalpha.com/input/?i=Integrate [Integrate[Sin[r^2]r%2C+{theta%2C+0%2C+pi}]%2C+{r%2C0%2C+4}] check it out again

15. experimentX

Integrate[Integrate[Sin[r^2]r, {theta, 0, pi}], {r,0, 4}]

16. anonymous

hw?

17. anonymous

i need guidance pls...

18. experimentX

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19. experimentX

do your integration and arrive at $\pi [ \cos 0 - \cos 16] \over 2\\ = {\pi(1 - \cos 16) \over 2 }$just use half angle formula

20. anonymous

do i need to use mehod by substitution for r^2