I am learning multivariable calculus myself and came across the double integral $\iint_R x^2\, \mathrm dA$ where $R$ is the region bounded by $y=x,y=\frac{16}x,x=8,y=0 $.
I was able to evaluate the integral by treating the region as a type I region.
I got $$ \iint_R x^2\,\mathrm dA = \int_0^4\int_0^x x^2\,dydx +\int_4^8 \int\limits_0^{16/x}x^2\, dydx = 448$$
But I wasn't able to evaluate it by treating the region as a type II region, as I can't set the limits.
- Is my result correct?
- How can I set the limits for the type II region?
You want to change the order of integral from $dy \ dx$ to $dx \ dy$.
At intersection of $xy = 16$ and $y = x$, $x = y = 4$
At intersection of $x = 8, xy = 16$, $y = 2$.
For $0 \leq y \leq 2$, region is bound between lines $y = x$ and $x = 8$.
For $2 \leq y \leq 4$, region is bound between line $y = x$ and curve $xy = 16$.
Please see the shaded area in the diagram. That is the region you need to integrate over.
The integral will be,
$\displaystyle \int_0^2 \int_y^8 x^2 \ dx \ dy + \int_2^4 \int_y^{16/y} x^2 \ dx \ dy$