Question

user polar coordinates to evaluate the integral of region of (x+y)dA, region r is bounded by the graph of x^2+y^2 = 2y

Answer 1

http://www.wolframalpha.com/input/?i=x%5E2%2By%5E2+%3D+y

Answer 2

x+y is a plane

Answer 3

i just don't get how to get the limits =\

Answer 4

radius ... 0.5

Answer 5

na the limit is 2sintheta =\

Answer 6

\[ \int_{0}^{1} \int_{0 }^{+ \pi} f(r, \theta) rd\theta dr\]
By looking at graph!!

Answer 7

but that's not right =\
\[\int\limits_{0}^{pi/2}\int\limits_{0}^{2\sin \theta} (r \cos \theta + rsin \theta) r dr d \theta\]

Answer 8

Oo ... sorry!! I think I made mistake!! r is dependent on theta

Answer 9

is the answer, but i have no idea how she go that

Answer 10

looks like today's not my day!!
http://www.wolframalpha.com/input/?i=r+%3D+2+sin+theta
\[ \int_{0}^{\pi}\int_{0}^{2\sin\theta} f(r,\theta)rdrd\theta\]

Answer 11

If you look at the graph ... theta goes from 0 to pi
and ... r varies as 2sin\theta, so limits for r should be 0 to 2sin\theta

and also integrate r first!!

now sure about pi/2

Answer 12

what if i don't have the graph?

Answer 13

i know i'm supposed to use what x and y equal to in polar coordinates, so like
x=rcostheta
y=rsintheta
x^2+y^2=r^2, but i don't understand how to do this in this instance..

Answer 14

wait nvm i think i figured it out haha

Answer 15

??
graphing your boundary is important ... it's the first thing you must do before you integrate!!

dxdy = r dr d\theta

Answer 16

f(x,y) = x + y => r(cos \theta + sin\theta)

Answer 17

what if i don't have a graphing calculator?

Answer 18

oh you were right i was looking at the wrong problem, it's o to pi ^.^

Answer 19

but yeah i don't know how to graph it without a calculator =\

Answer 20

you are supposed to do it by hand ... graphing curves are taught before or at single variable calculus ...!!

Answer 21

i def don't remember that lol i took it four years ago :(

Answer 22

http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/lecture-10-curve-sketching/

Answer 23

lol my final is tomorrow night, i can't spend an hour listening to a lecture =\

Answer 24

i'm just going to assume that since its x and y squared, it's a circle, and the 2y means that i moves up on the y axis. anyway, how do i integrate this? :(
\[\int\limits_{0}^{pi}\sin^3 \theta \cos \theta + \sin^4 \theta d \theta\]

Answer 25

integrate it individually ...
use substitution ...
\[ \int \sin^3x \cos x dx = ... \]

change sin^4x = (sin^2x)^2 = ((1- cos 2x)/2)^2 ... square it ... again change cos^2 2x ... using double formula ...
LOL ... it's going to long ...

Answer 26

the later part was for sin^4x

Answer 27

i don't even know what you just did lol

Answer 28

sorry ... i didn't get notification

Answer 29

http://www.wolframalpha.com/input/?i=integrate+sin%5E3x+cos+x
click show steps ... to view steps

Answer 30

for the other
http://www.wolframalpha.com/input/?i=integrate+sin%5E4x