# Area of the unit circle using integrals

In Calculus by Michael Spivak, the area of the unit circle is explored using integrals:

If $0 \le x \le 1$, this area can be expressed as the sum of the area of a triangle and the area of a region under the unit circle:

$\dfrac{x\sqrt{1 - x^2}}{2} + \int_{x}^{1} (\sqrt{1 - t^2}) dt$

I am struggling to understand the expression $\dfrac{x\sqrt{1 - x^2}}{2} + \int_{x}^{1} (1 - t^2) dt$.

As I understand it, the term $\sqrt{1 - x^2}$ is a semicircle, and the term $\dfrac{x}{2}$ in $\dfrac{x\sqrt{1 - x^2}}{2}$ is the area of a sector of the unit circle. I also understand the $\int_{x}^{1} (\sqrt{1 - t^2}) dt$ term.

But why do we have the term $\dfrac{x\sqrt{1 - x^2}}{2}$? How does it contribute to, as the author says, finding the area of a unit circle if $0 \le x \le 1$?

I would greatly appreciate it if people could please take the time to break this expression down for me and clarify its components.

• I feel like these is some missing context here. But if it's of any help, then $\frac{1}{2}\cdot x \cdot \sqrt{1-x^2}$ is the area of the triangle such that one of the legs goes from the origin to the $(x,0)$ and the other leg extends from $(x,0)$ and up to the unit circle. – benguin Mar 4 '17 at 14:38
• @benguin thanks for the response. Can you please elaborate on how $\frac{1}{2}\cdot x \cdot \sqrt{1-x^2}$ is the area of a triangle? I understand that area of a triangle $= \dfrac{1}{2} * base * height$, but how is $\sqrt{1-x^2}$ height? As we know, $\sqrt{1-x^2}$ is just the equation of a semicircle on the positive y-axis? – The Pointer Mar 4 '17 at 15:16
• Yes, exactly, if you take the equation $y = \sqrt{1-x^2}$ and interpret $y$ as "the height" (the height from the x-axis to be specific), then you quite literally have that "the height is $\sqrt{1-x^2}$". – benguin Mar 4 '17 at 15:19

"This area" is seen here: It is a sector of a unit circle, the top corner has a given $x$ coordinate.
The unit circle has equation $x^2+y^2=1$. The top half of the circle has equation $y = \sqrt{1-x^2}$.
The green triangle has area $$\frac{x\sqrt{1-x^2}}{2}$$ (base times height divided by $2$). And the blue region has area $$\int_x^1 \sqrt{1-t^2}\;dt$$