# Finding the area of shaded region

Consider a circle with the equation $$x^2 + y^2 = 1$$. Now find the area of the shaded region which we denote by $$u$$.

I'm trying to approach this question and would like support on my integral calculation. The difficulty I'm having is with substitution.

For example, given that:

$$u = (OM)(MP)+2\int^{1}_{x} \sqrt{1-x^2} \space dx$$

I should be getting:

$$u = xy + [x\sqrt{1-x^2}-\cos^{-1}x]^{1}_{x}$$

Although, I currently lack the technical skills to get this answer.

My approach:

I thought that I could substitute back in $$x$$ for $$\sin(x)$$ to get $$\sqrt{1-\sin^2(x)}$$, then when $$u = \sin(x) \implies du = \cos(x)$$, so that I get:

$$2\int^{1}_{x} \sqrt{1-\sin^2(x)}\cdot \cos(x) \space dx$$

However, this does not lead me close to the answer.

I have also tried inputting this equation into symbolab, and get:

$$\arcsin \left(x\right)+\frac{1}{2}\sin \left(2\arcsin \left(x\right)\right)+2C$$

Which is also wrong. What might be the approach towards this?

In Progress:

I've currently rearranged my substitution and it looks promising:

$$u = 1-\sin^2 (x) \implies du = -cos(x)dx \implies-arcos(x)du = dx$$

I cannot manage with this approach neither, as I'll have to integrate the square root.

Looking at the equation, I'll need something like:

$$secant(x) -arcos(x)$$ to convert the secant into $$x\sqrt{x^2-1}$$

• please do not write $x = \sin x$. Use a different variable. It is confusing otherwise and you are likely to make mistakes. – Math Lover Feb 28 at 13:29
• @MathLover I corrected it. I can see how it can lead it mistakes, and I'll be sure to correct my approach from now on. Thanks for the tip. – Meilton Feb 28 at 13:37

## 3 Answers

You may integrate by parts directly

\begin{align} \int_{x}^{1}\sqrt{1-t^2}dt & =\int_{x}^{1}\frac{\sqrt{1-t^2}}{2t}d(t^2) = -\frac12 x\sqrt{1-x^2}+\frac12 \cos^{-1}x \end{align}

which, with $$y=\sqrt{1-x^2}$$, leads to the area $$u =xy + 2 \int_{x}^{1}\sqrt{1-t^2}dt =\cos^{-1}x$$

Alternative method:

The area of the shaded region is $$\frac{1}{2}r^2\theta,\$$ where $$r$$ is the radius of the circle and $$\theta\$$ is the angle in radians that the region makes at the centre of the circle $$O$$.

$$r = 1,\$$ therefore the area is $$\frac{1}{2}\times1^2\times \theta = \frac{1}{2}\theta.$$

Now, $$\angle POM = \frac{\theta}{2},\$$ so we can express $$\frac{\theta}{2}\$$ in terms of the $$x-$$coordinate of $$M = (M_x,M_y)$$, $$M_x$$ say. Looking at the triangle $$OPM,\$$ we know that $$OP=1,\$$ therefore $$\frac{\theta}{2} = \cos^{-1}\left(\frac{M_x}{1}\right) = \cos^{-1}(M_x) \implies \theta = 2\cos^{-1}(M_x).$$

So area of shaded region $$= \frac{1}{2} \times 2\cos^{-1}(M_x) = \cos^{-1}(M_x).$$

(I assume that $$(x,y)$$ is in the first open quadrant).

The area of a circular sector with angle $$\alpha$$ is

$${\frak A}=\alpha/2$$

(in particular, if $$\alpha=2\pi$$, we get an area equal to $$\pi$$ in conformity with formula $$\pi R^2$$ for $$R=1$$).

As, in triangle $$OMP$$,

$$\alpha/2=\operatorname{atan}\dfrac{MP}{MO}=\operatorname{atan}\dfrac{y}{x}$$

As a consequence:

$${\frak A}\text{=}\operatorname{atan}\dfrac{y}{x}$$