Finding the Area of a Disc with Integration Methods

[markosheehan]

Use integration methods to establish the formula A = π r^2 for the area of a disc ofradius r.

so the equation of the circle is x^2 +y^2 =r^2 . i will try and find the area of one quadrant using integration. so it will be ∫ r,0 y dx
so ∫ r,0 √(r^2 -x^2) dx so from here i am trying to integrate it but it is not working for me. i saw some suggested help saying you should substitute x for something but i do not know what it is and i can not see why you would do this either. i can't see why my normal method should not work.

 

[MarkFL]

Okay, so what you want using this approach is:

\(\displaystyle A=4\int_0^r \sqrt{r^2-x^2}\,dx\)

We need to multiply by 4 since we are integrating over just 1 of the 4 quadrants, that is the integral is 1/4 of the total area of the disc. For an integral of this type, a good substitution is:

\(\displaystyle x=r\sin(\theta)\implies dx=r\cos(\theta)\,d\theta\)

So now, we want to change the limits, the integrand and the differential from being in terms of $x$ to being in terms of $\theta$. This will give us:

\(\displaystyle A=4\int_0^{\frac{\pi}{2}} \sqrt{r^2-r^2\sin^2(\theta)}r\cos(\theta)\,d\theta\)

In the integrand, we can factor the radicand and then pull any constants out front:

\(\displaystyle A=4r^2\int_0^{\frac{\pi}{2}} \sqrt{1-\sin^2(\theta)}\cos(\theta)\,d\theta\)

Can you use a Pythagorean identity to simplify the radical?

 

[markosheehan]

x=rsin(θ) ⟹dx=rcos(θ)dθ

I am not sure how you get this. Where does the d and the dθ come from.? Are you not just supposed to integrate both sides?

I can take it from where you left it though.

 

[MarkFL]

x=rsin(θ) ⟹dx=rcos(θ)dθ

I am not sure how you get this. Where does the d and the dθ come from.? Are you not just supposed to integrate both sides?

I can take it from where you left it though.

This is just a substitution:

\(\displaystyle x=r\sin(\theta)\)

And via differentiation, we obtain:

\(\displaystyle dx=r\cos(\theta)\,d\theta\)

This takes care of the integrand and the differential, and for the limits, we observe that our substitution may be written:

\(\displaystyle \theta(x)=\arcsin\left(\frac{x}{r}\right)\)

And we have:

\(\displaystyle \theta(0)=\arcsin\left(\frac{0}{r}\right)=0\)

\(\displaystyle \theta(r)=\arcsin\left(\frac{r}{r}\right)=\frac{\pi}{2}\)

 

[MarkFL]

A simpler approach would be to add via integration all of the circumferences of the concentric circles from radius 0 to radius r:

\(\displaystyle A=\int_0^r (2\pi u)\,du=2\pi\int_0^r u\,du\)

 

[markosheehan]

This is just a substitution:

\(\displaystyle x=r\sin(\theta)\)

And via differentiation, we obtain:

\(\displaystyle dx=r\cos(\theta)\,d\theta\)

This takes care of the integrand and the differential, and for the limits, we observe that our substitution may be written:

\(\displaystyle \theta(x)=\arcsin\left(\frac{x}{r}\right)\)

And we have:

\(\displaystyle \theta(0)=\arcsin\left(\frac{0}{r}\right)=0\)

\(\displaystyle \theta(r)=\arcsin\left(\frac{r}{r}\right)=\frac{\pi}{2}\)

i am sorry but i have never done this before and i am confused. thanks anyway

 

[MarkFL]

i am sorry but i have never done this before and i am confused. thanks anyway

Let me know where the confusion lies, and I'll try to explain. :)