Question

find the area of the region that lies inside both curves r=2sintheta r=2costheta

Answer 1

is this the graph?

Answer 2

yes im having trouble find the point of intercepts too

Answer 3

2sintheta=2costheta

Answer 4

you just need to find the area of half of it; and then double that

they intersect along the line y=x; or theta = 45

Answer 5

\[\int_{0}^{pi/4}\int_{0}^{r(t)} r\ drdt\]

Answer 6

id use r = 2cos(t) for my upper radial limit

Answer 7

\[\int_{0}^{pi/4}\int_{0}^{2cos(t)} r\ drdt\]
\[\int_{0}^{pi/4}\frac{1}{2}(2cos(t))^2dt\]

\[\int_{0}^{pi/4}2cos^2(t)dt\]

Answer 8

then id double that result

Answer 9

lol, um ... 2sin(t) is the radial limit of 0 to pi/4 tho so maybe just change that little bit

Answer 10

\[2\int_{0}^{pi/4}2sin^2(t)dt\]
thats better

Answer 11

ok i got it thanks