We are asked to find the area between the circle $x^2 + y^2 = 4$ and the line $x + y = 2$ lying in the first quadrant.
One method is to find the area by integrating $2-x$ from $0$ to $2$ and subtracting that from $\sqrt{4-x^2}$ from $0$ to $2$
which gives the answer $\pi – 2$.
If I had to solve this same equation using double integrals, can someone explain the logic behind taking $1$ as the function which is integrating over $y = 2-x$ to $y= \sqrt{4-x^2}$ and $x = 0$ to $x = 2$
I'm aware this is a very basic question but I haven't understood it, sorry.
Best Answer
You can start from $dA=dxdy$. To compute the area, you need to find $$A=\int_Ddx\,dy,$$ where $D$ is the region you are interested in. Now let's try to describe this region $D$. You can cut this region into vertical slices. When $x$ is fixed, $y$ will grow from $2-x$ to $\sqrt{4-x^2}$. So the "height" of the region at this point $x$ is $\int_{2-x}^{\sqrt{4-x^2}}\,dx$. Then you can integrate over $x$. So you will have \begin{align*} A&=\int_D\,dx\,dy\\ &=\int_0^2\left(\int_{2-x}^{\sqrt{4-x^2}}\,dy\right)\,dx\\ &=\int_0^2\left[\sqrt{4-x^2}-(2-x)\right]\,dx. \end{align*} This can be solved by 1d-calculus.