Integration area question (1 Viewer)

chingyloke

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-Find the area bounded by the curve y = (square root of) 4-x^2, the x-axis and the y-axis in the first quadrant.

Now it forms a semicircle with x-intercepts -2 and 2.

So the area can be found by

A=pi r^2/4
= pi x 2^2/4
=pi

but can it be found by using the traditional finding the primitive and subbing in the limits? in this case 0 and 2?

i can't get it that way...not sure if i'm stuffing up the algebra or what...
 

Timothy.Siu

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yeah i just did it, but i might not be fully correct, we haven't learnt this yet.

soo
2
Sroot(4-x^2)dx
0

let x=2sin u
dx=2cos u du

x=0 u=0
x=2 u=pi/2

sub in x=2sin u and dx=2cos udu
pi/2
S4cos^2u du
0

cos 2u=2cos^2 u-1
4cos^2 u=2cos 2u+2
pi/2
S (2cos 2u+2)du=[sin 2u+2u]from pi/2 to 0 =0+pi=pi
0
 
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Valupatitta

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whats the answer?
i got 17 1/15.
if its right then u integrate normally, except the limits are x=2 and x=0 because its in the first quadrant.
 
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chingyloke

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Valupatitta said:
whats the answer?
i got 17 1/5.
if its right then u integrate normally, except the limits are x=2 and x=0 because its in the first quadrant.
The answer is pi...so approx. 3.141 units^2
 

tommykins

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chingyloke said:
-Find the area bounded by the curve y = (square root of) 4-x^2, the x-axis and the y-axis in the first quadrant.

Now it forms a semicircle with x-intercepts -2 and 2.

So the area can be found by

A=pi r^2/4
= pi x 2^2/4
=pi

but can it be found by using the traditional finding the primitive and subbing in the limits? in this case 0 and 2?

i can't get it that way...not sure if i'm stuffing up the algebra or what...
you need 4unit trig substiution to do this question without the semi circle identification.
 

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