Question

on xy plane(real plane) let S(x1) be the area of overlap of two region defined by following formulas x^2+y^2<=a^2 & 0<=x<=x1 where x1<=a,find S(x1)..plzzzzzzzz help me in this problem,thankyou

Answer 1

Use a TI-Nspire calc to figure it out

Answer 2

i want to solve it on paper and to know the procedure

Answer 3

x^2 + y^2 = a^2 is a circle centred at (0,0) of radius a.

Your inequality I'm guessing is intended to define a line segment on the x-axis of maximum length equal to the radius.

That answer your question?

Answer 4

\[y = \sqrt{a^2 -x^2}\]
area = \[\int\limits_{0}^{x_1} y dx\]

Answer 5

That's a good answer if a variable area is what op is looking for, I don't think the question was clear.

Answer 6

actually i guess it would be twice the integral
\[2\int _0^{x_1} \sqrt{a^2-t^2}dt\]

Answer 7

so now you have to find the integral. i would look in the back cover of the calc text

Answer 8

wow almost the first one. says
\[\int\sqrt{a^2-u^2}du=\frac{u}{2}\sqrt{a^2-u^2}+\frac{a^2}{2}\sin^{-1}(\frac{u}{2})\]

Answer 9

so multiply by 2, plug in
\[u=x_1\] and
\[u=0\] and subtract. easy especially since 0 gives you 0. so your answer is right in front of you without the 2 in the denominator and with u replaced by
\[x_1\]

Answer 10

the question was to find S(x1) & d/d(x1) of S(x1)...anyway thanks for the help,i think this much of ur help is really appreciable,atleast u gave me an insight :)

Answer 11

hold the phone. isn't that the answer? S(x1) is just that integral

Answer 12

yea

Answer 13

so answer is
\[S(x_1)=x_1\sqrt{a^2-x_1^2}+a^2\sin^{-1}(\frac{x_1}{2})\]

Answer 14

thankyou dear frnd :)..appreciate ur kindness

Answer 15

yw hope it is clear

Answer 16

btw the derivative of the integral is the integrand so that question is answered as well