# Find the area

1. Oct 8, 2004

### .....

i made a pretty picture but i cant get it to upload

oh well...i'll try to explain it

a circle of radius 3cm is inside a circle of radius 8cm (in the middle). On top of these two circles is a wedge (i think...it's like a triangle except curved at the end to fit the edge of the larger circle) which has an angle of 45 degrees at the centre of the cirlces.

what is the area of the larger circle without the wedge or the smaller cirlce?

I got an answer...but i'd like to see if it's right or not, and also if there's a simpler way of doing it.

This is what I did:

area of the bigger cirlce = 64pi
area of the smaller circle = 9pi

area of the larger wedge = integral of x - (64 - x^2)^1/2 between 0 and 8

i havent actually learned how to do that algebraicly, but there's this nifty little function on my calculator...
i get 50.27cm squared

(can someone show me how to integrate that function anyway just for the hell of it?)

area of the smaller wedge = integral of x - (64 - x^2)^1/2 between 0 and 3
= 23.42cm squared

(64pi - 9pi) - (50.27 - 23.42) = 145.95cm squared

so...

Q1. Did I do it right?
Q2. If not, how do i do it?
Q3. If so, is there an easier way i could have done it?
Q4. How do you integrate non-linear functions raised to a power?

2. Oct 8, 2004

### .....

area of the smaller wedge = integral of x - (9 - x^2)^1/2 between 0 and 3
= 2.57cm squared that shouldve been...

but it's wrong anyway....

why?

if the middle of the circles was the origin, take one of the lines formed by the wedge as the x axis, then the other line is y = x because of the 45 degree angle. the curved line at the end was part of the cirlce, so it had the equation y = (9 - x^2)^1/2

so then why shouldnt the integral of x - (9-x^2)^1/2 between 0 and 3 be the area of the smaller wedge?

3. Oct 8, 2004

### HallsofIvy

The area of a "wedge" of angle &theta; (measured in radians) in a circle of Radius R is &theta R2/2. That's just the area of the circle &pi;R2 time the fraction &theta;/2&pi;.

Since your larger circle as radius 8 cm, a 45 degree= &pi;/4 radian angle cuts out a wedge of area (&pi;/8)(64)= 8&pi; square cm.

In the smaller circle, of radius 2 cm., that same wedge cuts out an area of (&pi;/8)(4)= &pi;/2 square cm. We need to know that since we don't want to subtract that area twice!

The entire circle of radius 2 cm has area 4&pi; square cm so the area of the small circle NOT in the wedge is (4- 1/2)&pi;= 7&pi;/2 square cm.

The large circle has area 64&pi; square cm. so the area left after removing the wedge and the circle (an upside down keyhole?) is 64&pi;- 8&pi;- 7&pi/2= 105&pi;/2 square centimeters.