## Help with Reimann(?) Integral

Say we have a circle of unit radius centered on the origin, and another point on the x-axis d units away, were we to draw a line at angle θ relative to the x axis, cutting across the circle at either 2, 1 or 0 points, the length of the chord made(or not made) is

$L(\theta) = 2r\sqrt{1-\frac{d^2\sin^2(\theta)}{r^2}}$

My calculus isn't up to par to figure out how to then determine the area of the circle using that equation...I would have thought maybe using the area of sector .5r^2dθ with r = L(θ) but I keep getting the wrong answers, is it not possible to integrate over theta here and get the area of the circle? Is the above function incorrect?

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Hi Zula110100100!
 Quote by Zula110100100 to then determine the area of the circle using that equation...… I would have thought maybe using the area of sector .5r2dθ with r = L(θ) but I keep getting the wrong answers
Your formulas look ok, and integrating from θ = 0 to π should work …
show us what you did

 I put it all together to get: $A = \int_{-\arcsin(\frac{r}{d})}^{\arcsin(\frac{r}{d})}{\frac{1}{2}(4r^2-4d^2\sin^2(\theta))}d\theta = 2r^2\theta + d^2\sin(2\theta)-2d^2\theta |_{-\arcsin(\frac{r}{d})}^{\arcsin(\frac{r}{d})}$ $A = 2\sin^{-1}(\frac{1}{1.5}) + (1.5)^{2}\sin(2\sin^{-1}(\frac{1}{1.5}))-2(1.5)^{2}\sin^{-1}(\frac{1}{1.5}) - (-2\sin^{-1}(\frac{1}{1.5}) - (1.5)^{2}\sin(2\sin^{-1}(\frac{1}{1.5}))+2(1.5)^{2}\sin^{-1}(\frac{1}{1.5})) = 4\sin^{-1}(\frac{1}{1.5}) +2(1.5)^{2}\sin(2\sin^{-1}(\frac{1}{1.5}))-4(1.5)^{2}\sin^{-1}(\frac{1}{1.5}) \approx -.65$ The domain of L(θ) 1-d^2sin^2(θ)/r^2≥0 d^2sin^2(θ)/r^2≤1 sin^2(θ) ≤ r^2/d^2 sin(θ) ≤ r/d θ <= arcsin(r/d)

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## Help with Reimann(?) Integral

Hi Zula110100100!

(just got up )
 Quote by Zula110100100 The domain of L(θ) 1-d^2sin^2(θ)/r^2≥0 d^2sin^2(θ)/r^2≤1 sin^2(θ) ≤ r^2/d^2 sin(θ) ≤ r/d θ <= arcsin(r/d)
but r > d, so r/d > 1, so arcsin doesn't exist!

the limits are 0 to π …

the pivot point (at d) is fixed, and the line rotates from horizontal to horizontal
 I put it all together to get: $A = \int_{-\arcsin(\frac{r}{d})}^{\arcsin(\frac{r}{d})}{\frac{1}{2}(4r^2-4d^2\sin^2(\theta))}d\theta = 2r^2\theta + d^2\sin(2\theta)-2d^2\theta |_{-\arcsin(\frac{r}{d})}^{\arcsin(\frac{r}{d})}$ $A = 2\sin^{-1}(\frac{1}{1.5}) + (1.5)^{2}\sin(2\sin^{-1}(\frac{1}{1.5}))-2(1.5)^{2}\sin^{-1}(\frac{1}{1.5}) - (-2\sin^{-1}(\frac{1}{1.5}) - (1.5)^{2}\sin(2\sin^{-1}(\frac{1}{1.5}))+2(1.5)^{2}\sin^{-1}(\frac{1}{1.5})) = 4\sin^{-1}(\frac{1}{1.5}) +2(1.5)^{2}\sin(2\sin^{-1}(\frac{1}{1.5}))-4(1.5)^{2}\sin^{-1}(\frac{1}{1.5}) \approx -.65$
no, the integrand has a square-root

however, although i said that you could solve it that way, on second thoughts, i don't think that √(r2 - d2sin2θ) has a straightforward integral
is this a question from a book, or one you've made up out of interest?

 out of interests...d>r, (d,0) is a point outside the circle. So arcsin(r/d) should I think somehow be the angle, relative to the x-axis, that a tangent line to the circle passes through (d,0) Does the root not go away from the fact that I am using L(θ) as r in .5r^2θ so .5(L(θ))^2θ and the root gets squared to find the area of each sector
 So here is a graph on a unit circle centered at origin, y=sqrt(1-x^2) and a line that passes through (d,0), d=1.5, the slope of that line is -tan(arcsin(r/d))=-tan(arcsin(1/1.5)), as you can see it is the correct slope, so greater than that there are no points of intersection, and less than that and greater than it's negative(consider +/- sqrt circle) there are two points of intersection, L(θ) from above should give the distance between those two points of intersection, and thus a chord from the circle. I am attemepting to integrate by adding areas of sectors that are between two such chords, though that area is not truly a sector, or even too very close to one...but, I was under the impression that as dθ approached 0 it would be close enough...more importantly though, why does the sqrt not go away when being squared? I am not sure how to shade the sector with the prgrm I am using, but it would be between the two lines.
 Just realized I didn't mention d>r in the first post and I really should have, does the formula still look correct for that case?
 Using the program to get the graphs there i put in the forumla for L(θ) and it has an option to integrate for area and it gave me 2.23...so my question is why can't I integrate for area this way? the distance from the point to the region being measured makes it impossible to use theta to integrate?

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