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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.

Mathematics
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as for radius is from -4 to 4. for angle is from 0 to 180 ?
yep!!
differentiate by r dr dθ ?

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Other answers:

use radiant (0 to pi)!! and dA = r dr d(theta)
ok. let me work out....
\[ \int_0^4 \int_0^\pi \sin ( r^2 \cos^2 \theta + r^2 \sin^2 \theta) \; r \; dr\; d\theta \]
could u pls explain to me why e radius is from 0 to 4 nt -4 to 4? and hw to get e function?
ok i understood. hw abt e integration part?
\[ \int_0^4 \int_0^\pi \sin ( r^2) \; r \; d\theta\; dr\ \\ \int_0^4 \sin ( r^2) \; r \; \left [ \theta \right ]_0^\pi dr\\ \] just use subs r^2 = u
ok...thanks. let me work out
is the ans \[- \pi \cos8
\[-\pi \cos 8\]
am i right??
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
Integrate[Integrate[Sin[r^2]r, {theta, 0, pi}], {r,0, 4}]
hw?
i need guidance pls...
|dw:1353425265831:dw|
do your integration and arrive at \[ \pi [ \cos 0 - \cos 16] \over 2\\ = {\pi(1 - \cos 16) \over 2 }\]just use half angle formula
do i need to use mehod by substitution for r^2

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