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

Question

amistre64 · Answer

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is this the graph?

anonymous · Answer

yes im having trouble find the point of intercepts too

anonymous · Answer

2sintheta=2costheta

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

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

amistre64 · Answer

then id double that result

amistre64 · Answer

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

amistre64 · Answer

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

anonymous · Answer

ok i got it thanks