Area between two curves question that doesn't involve calculus (1 Viewer)

[Dota55]

Find the exact area enclosed between the the curve y = square root(4-x^2) and x - y + 2 = 0


The answer is (Pi - 2) units squared.....and i can't see why..

Can someone please explain??

 

[tacogym27101990]

yeah this doesnt invole any integration.

y=sqr(4-x^2) is a semi circle of radius 2
but x-y+2=0 is intersects with the semi circle at x=0 and x=-2

so what you actually want to do, is find the area of the quarter circle, and the area of the triangle made by x-y+2=0 and the x axis and y axis. then minus the triangle from the semi circle

so we have: (pi*r^2)/4 - .5*2*2 r=2
= 4pi/4 - 2
= pi - 2 u^2

hope it makes a bit of sense, would be easier if i knew how to put a graph in here

 

[Dota55]

hey wait, hold on.

Yeah this is my problem: Isn't it Pi + 2???

I mean, we're adding the area of the quarter circle and the triangle in the 2nd quadrant (and its area, hence there's no negative values)

Thats my confusion, i got pi + 2 when i first did it.

 

[cwag]

No, to find the area we subtract the minor area away from major area...pi-2. this gives a postive answer....alternatively, i don't know if this is done in 2U course, but the area of a minor segment is A= (1/2)r²(@-sin@) where @ is radians..
so we could work this out by..

A= (1/2)*4*(pi/2 - sin pi/2)
= pi - 2

 

[tacogym27101990]

but that ways a bit unessecary in a question like this isnt it?

 

[cwag]

well, no, it didn't specify in which way to answer the question. So we are permitted to use any method we like. Personally, i find it easier, and quicker...but it doesn't matter, whatever suits u, or comes to u at the time

 

[addikaye03]

This question is from the Maths Focus book isn't it? i thought it seemed familiar lol

 

[conics2008]

when you say calculus ok..

use area of circle /2 and that line looks like a triangle.. and minus it =)