Question

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.

Answer 1

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

Answer 2

yep!!

Answer 3

differentiate by r dr dθ ?

Answer 4

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

Answer 5

ok. let me work out....

Answer 6

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

Answer 7

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

Answer 8

ok i understood. hw abt e integration part?

Answer 9

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

Answer 10

ok...thanks. let me work out

Answer 11

is the ans \[- \pi \cos8

Answer 12

\[-\pi \cos 8\]

Answer 13

am i right??

Answer 14

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

Answer 15

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

Answer 16

hw?

Answer 17

i need guidance pls...