Question

Sketch the region $R$ whose area is given by the double integral. Then change the order of integration and show that both orders yield the same area. $${\int }_{-2}^2{\int }_0^{4-y^2}dxdy$$

Short answer

The area computed by both integrals is the same and equal to $16/3$. The region $R$ is a semi-circle with radius 2 centered at the origin in the xy-plane.

Step-by-step solution

Identify the region and sketch it

Identify the region $R$ from the given limits of integration: $x$ ranges from $0$ to $4-y^2$ (which is a semi-circle), and $y$ ranges from $-2$ to $2$. So, the region $R$ can be represented as a semi-circle with radius 2 and centered at the origin in the xy-plane.

Change the order of integration

The goal now is to express $y$ as a function of $x$. Observing the semi-circle, we note that given a certain $x$, $y$ ranges from $-\sqrt{4-x^2}$ to $\sqrt{4-x^2}$, whereas $x$ ranges from $0$ to $2$. Thus, the double integral can be rewritten as $${\int }_0^2{\int }_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dydx$$

Compute the area using both orders of integration

First compute the original integral $${\int }_{-2}^2{\int }_0^{4-y^2}dxdy={\int }_{-2}^2[x{]}_0^{4-y^2}dy={\int }_{-2}^2(4-y^2)dy=[4y-\frac{y^3}{3}{]}_{-2}^2=16/3$$ Then compute the integral with the order of integration reversed $${\int }_0^2{\int }_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dydx={\int }_0^2[y{]}_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}dx={\int }_0^{2}2\sqrt{4-x^2}dx=\frac{1}{2}\pi r^2=16/3$$ Thus, both orders of integration yield the same area.