$\displaystyle{\int \int 1 dxdy }$ over the region $R$ where $R = \{(x,y):x^2+y^2=16\}$
The solution given in class went like this:
$\displaystyle{\int_{-4}^{4} \int_{x=-\sqrt{4-y^2}}^{x=\sqrt{4-y^2}}1dxdy} = 2\int_{0}^{4}\int_{x=0}^{x=\sqrt{4-y^2}}dxdy$
The other question went like this
$\displaystyle{\int \int (x^2+y^2-1) dxdy} $ over the region $R$ where $R = \{(x,y):x^2+y^2=16\}$
The solution for this problem went like this:
$\displaystyle{\int_{-4}^{4} \int_{x=-\sqrt{4-y^2}}^{x=\sqrt{4-y^2}}(x^2+y^2-1)dxdy = 4\int_{0}^{4}\int_{x=0}^{x=\sqrt{4-y^2}}(x^2+y^2-1)dxdy}$
Is this a typo or there is a logic behind this?
Notice that the integrand is even for the component $x$. That is, $$f(-x,y)=f(x,y).$$ With that you can justify converting the integral $\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}dx$ into $2\int^{\sqrt{4-y^2}}_{0}dx$.
On the other hand, the function $\int f(x,y)dx$ is even for the component $y$. That is, $$\int f(x,-y)dx=\int f(x,y)dx.$$ This justifies splitting the other integral, $\int_{-4}^4$.