Definite integral of a quarter circle

[Painguy]

Homework Statement

Find the definite integral of a quarter circle.

Homework Equations

x^2 +y^2=10

The Attempt at a Solution

x^2=10-y^2
x=sqrt(10-y^2)

∫ sqrt(10-y^2)dy from 0 to sqrt(10)

I'm not sure what to do here.

 

[Mute]

Well, it's a circle, so you can probably guess that a factor of $\pi$ should pop out somehow. What kind of substitution do you think that fact might suggest you should use?

 

[Painguy]

I guess I should use a trig substitution. 10cos(θ)^2 +10sin(θ)^2=10
√(10) cos(θ) = √(10-10sin(θ)
y=√(10)sin(θ)
∫√(10-(10)sin(θ)) dy
dy=√(10)cos(θ)dθ
∫√(10-(10)sin(θ)) √(10) cos(θ) dθ
∫√(10) cos(θ) √(10) cos(θ) dθ
arcsin(t/sqrt(10))=θ
10 ∫cos(θ)^2 dθ

how do i evaluate that last integral?

 

[Mute]

I guess I should use a trig substitution. 10cos(θ)^2 +10sin(θ)^2=10
√(10) cos(θ) = √(10-10sin(θ)
y=√(10)sin(θ)
∫√(10-(10)sin(θ)) dy
dy=√(10)cos(θ)dθ
∫√(10-(10)sin(θ)) √(10) cos(θ) dθ
∫√(10) cos(θ) √(10) cos(θ) dθ
arcsin(t/sqrt(10))=θ
10 ∫cos(θ)^2 dθ

how do i evaluate that last integral?

There's an identity that relates $\cos^2\theta$ to $\cos(2\theta)$. Do you know it? If not, you can derive it. Consider the double angle formula for cos:

$$\cos(A+B) = \cos A \cos B - \sin A \sin B$$

Pick A and B appropriately and then use Pythagoras' theorem to derive the identity, then plug that into your integral.

 

[Painguy]

There's an identity that relates $\cos^2\theta$ to $\cos(2\theta)$. Do you know it? If not, you can derive it. Consider the double angle formula for cos:

$$\cos(A+B) = \cos A \cos B - \sin A \sin B$$

Pick A and B appropriately and then use Pythagoras' theorem to derive the identity, then plug that into your integral.

cos(2x)=cos^2(x)-1 +cos^2(x)
cos(2x)=2cos^2(x)-1
cos(2x)/2 +1/2

5∫ 1 dθ + ∫ cos(2θ) dθ

5θ + sin(2θ)/2 from 0 to pi/2

5pi/2 ?

Oh man that was not fun. :/ Thanks for your help. I REALLY appreciate it.

 

[Mute]

cos(2x)=cos^2(x)-1 +cos^2(x)
cos(2x)=2cos^2(x)-1
cos(2x)/2 +1/2

5∫ 1 dθ + ∫ cos(2θ) dθ

5θ + sin(2θ)/2 from 0 to pi/2

5pi/2 ?

Oh man that was not fun. :/ Thanks for your help. I REALLY appreciate it.

Yep, that's the answer! It's easy to check, too: since it's a quarter of a circle of radius $\sqrt{10}$, you expect the answer to be $\pi (\sqrt{10})^2/4 = 5\pi/4$, which is what you got.

There's another way to do this which is simpler, but you have to start it off a bit differently than you did, and I didn't know if you've studied polar coordinates or double integrals yet.

You integral is basically

$$\int_0^{\sqrt{10}} dy \int_0^{\sqrt{10-y^2}}dx;$$
doing the x integral gives you the integral you started with. However, converting to polar coordinates $x = r\cos\theta$, $y=r\sin\theta$, $dxdy \rightarrow d\theta dr r$ turns the integral into

$$\int_0^{\pi/2} d\theta \int_0^{\sqrt{10}} dr r,$$

which is much easier to do.

 

[HallsofIvy]

cos(2x)=cos^2(x)-1 +cos^2(x)
cos(2x)=2cos^2(x)-1
cos(2x)/2 +1/2

5∫ 1 dθ + ∫ cos(2θ) dθ

5θ + sin(2θ)/2 from 0 to pi/2

5pi/2 ?

Oh man that was not fun. :/ Thanks for your help. I REALLY appreciate it.

Odd. I would have considered that a lot of fun and very satisfying!