Integration area question (1 Viewer)

[chingyloke]

-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]

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

 

[Valupatitta]

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.

 

[chingyloke]

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

 

[chingyloke]

p.s tim your working is really intense.

 

[tommykins]

-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.