Prooving Area of Circle Equation

[rhyso88]

Hi, i am hoping that someone could help me or give me a hint for the follwing question...i will show what I have done so far

The area of a quarter circle given is

Integral sqrt(25-x^2) dx

Use the substitution x = 5 sinu to evaluate this def integral exactly an show that your answer is consistent with area pi*r^2 with radius r

firstly i did dx/du = 5cos u

then i said:
Integral 5sqrt(1-sin^2u).5cosu du
Integral 5cosu.5cosu
Integral 25 cos^2u du
Integral 25(1+cos2u)/2
Integral 25/2 +25cos2u/2

and well the problem i am having is that i don't know how to get the u =5sin u to SUB into the intergal so i can solve it (ie i can SUB u into cos2u...it doesnlt work)

so hopefully someone could point me in a better diretion to solve this??

thanks heaps!

 

[courtrigrad]

it shouldn't be $$x = 5\sin u$$, it should be $$x = 5\sin \theta$$.

 

[d_leet]

it shouldn't be $$x = 5\sin u$$, it should be $$x = 5\sin \theta$$.

And the difference between those two is?

 

[HallsofIvy]

and well the problem i am having is that i don't know how to get the u =5sin u to SUB into the intergal so i can solve it (ie i can SUB u into cos2u...it doesnlt work)

I don't understand what your question is. You seemed to be able to substitute and integrate!
If your question is "How do you know to make that substitution?", it's because of: sin²u+ cos²u= 1 so that
1- sin² u= cos²u, a "perfect square".

If your question is "how do I do that last integral", $\frac{25}{2}\int 1+ cos(2u) du$, you certainly should know that the integral of 1 du is u and the integral of cos(2u) is (1/2)sin(2u). That integral is
$\frax{25}{2}u+ \frac{1}{2}sin(2u)$. Since you say you are dealing with a quarter circle and you appear to have set it up with center at (0,0) and radius 5, your original integral is from x= 0 to x= 5. 5 sin u= x= 0 or sin u= 0 when u= 0 and 5 sin u= x= 5, or sin u= 1 when u= $\pi/2$ your limits of integration in the final integral are 0 and $\pi/2$.

 

[rhyso88]

sorry

the question is to proove using intergation techniques and this quarter circle as an example, that the area of a quarter circle is 1/4*pi*r^2. Ie intergrate the given integral and compare that to the area of a circle formula and see that they match. However, the question stipulates that you must use the substitution which I outlined...which leads to my prob when i get to the part where you substitute this : x = 5 sin u back into the equation which is meant to be easier to integrate... so i guess you could say

i don't know how to get u =5sin u (the upper limit) to substitute into the integral:

25/2 + 25 (cos 2u)/2 (which is as simplified as i can get the intergal)

thanks for all you responses btw! appreciated

 

[rhyso88]

No Its A Half Circle!

SORRY! just rechecked the Q, its a semi circle, not a quarter!

 

[rhyso88]

I understand the intergal you did completely, no probs there!

okay here's wat i don't get, how did you change the limits which i had i had, ie 0 and 5 sin u into numbers which I can actaually substitute into the intergal and then solve, it is irrlevant whether a quarter circle or half circle, but i chose quarter circle since the limits 0 and 5 seemed easier.

hang, i think i found my error, the limits therefore should be what you said, i was doing sumthin completely stupid with my limits, i don't know where 5sinu came from as a limit! sorry...THANKU HEAPS FOR YOU HELP!

 

[courtrigrad]

5 = 5sinu, so sin u = 1.
x^2 + y^2 = r^2, y^2 = sqrt(r^2 - x^2)