Question

\(\int\limits_0^2 {\int\limits_{{x^2}}^4 {f(x,y)dy} dx} \)

Short answer

We should know how to do graphical representation.

Step-by-step solution

Given data:

Consider the following double integral

\(\int\limits_0^2 {\int\limits_{{x^2}}^4 {f(x,y)dy} dx} \)

The objective is to sketch the region \(f\)integration and change the order of integration.

Graph:

Graphical representation

Region \(D\)can be expressed as which is of type-I

If \(f(x,y)\) is continuous on type-I region \(D\)such that \(D = \left\{ {(x,y)|a \le x \le b,{g_1}(x) \le y \le {g_2}(x)} \right\}\)

Graph:

Changing order of integration:

Region \(D\)scan be expressed as which is of type-II

If \(f(x,y)\) is continuous on type-I region \(D\)such that \(D = \left\{ {(x,y)|c \le y \le d,{h_1}(y) \le x \le {h_2}(y)} \right\}\)

The given region \(D = \left\{ {(x,y)|0 \le x \le 2,{x^2} \le y \le 4} \right\}\).

From the above figure the alternative description of \(D\)is \(y = {x^2}\) or\(x = \sqrt y \)

When \(y(x) = 4,x(y) = \sqrt y ,{\rm{ }}x(y) = 2,{\rm{ }}x(y) = 0,\) and \(x = 0,{\rm{ }}y = 0\).

When \(x = 2,{\rm{ }}y = 4\)

Now the region \(D\), \(D = \left\{ {(x,y)|0 \le x \le \sqrt y ,0 \le x \le 4} \right\}\)

So, the integral is \(\int\limits_{y = 0}^4 {\int\limits_{x = 0}^{\sqrt y } {f(x,y)dA} } \)

Hence, the change of order of integration is \(\int\limits_{y = 0}^4 {\int\limits_{x = 0}^{\sqrt y } {f(x,y)dA} } \).