Question

Explain the difference between a type I region and a type II region.

Short answer

It can be seen that, in the case of type I region, the integration is first done with respect to $y$ and then with respect to $x$.

Step-by-step solution

Given Information

Two regions are given. We need to find the difference between two.

Consideration

In interval $[a,b],a<b$, type I region is bounded by $y=g_1\left(x\right)$at the bottom and $y=g_2\left(x\right)$at top for all $x\in [a,b]$.

In interval$[c,d],c<d$, type I region is bounded by $x=h_1\left(y\right)$to the left and$x=h_2\left(y\right)$to the right for all$y\in [c,d]$

Simplification of first region 

The surface integral is calculated as ${\iint }_{\Omega }f(x,y)dA$

It can be calculated by taking an elemental area of breadth $\Delta x$, one end lying on $y=g_1\left(x\right)$and other on $y=g_2\left(x\right)$

Hence, surface integral is

${\iint }_{\Omega }f(x,y)dA={\int }_a^{b{g}_1}{\int }_{g_1\left(x\right)}^{g_2\left(x\right)}f(x,y)dydx$

Simplification of second region 

It can be calculated by taking an elemental area of breadth $\Delta y$one end lying on $x=h_1\left(y\right)$and other on role="math" localid="1653920549681" $x=h_2\left(y\right)$

The integral becomes

${\iint }_{\Omega }f(x,y)dA={\int }_{c{h}_4\left(y\right)}^{d{h}_2\left(y\right)}f(x,y)dxdy$