Question

Express D as a region of Type 1. And also as a region of type 2. Then evaluate the double integral in 2 ways.

D is enclosed by the lines \(y = x,y = 0,x = 1\)

Short answer

Therefore, answer is \(\frac{1}{3}\)

Step-by-step solution

It is easier to sketch the graph of the function

This particular region is both type 1 & 2.

\(\begin{array}{l}a \le x \le 6\\{g_1}(x) \le y \le {g_2}(x)\end{array}\)

Also written as:

\(\begin{array}{l}a \le x \le 6\\{g_1}(y) \le x \le {g_2}(y)\end{array}\)

Step 2:- Integrating on x.

And on graph, draw a line inside the region indicating direction of integration.

Step 3:- Now, observing that, \(0 \leqslant y \leqslant x\) and \(0 \leqslant x \leqslant 1\)

\(I = \int\limits_0^1 {\int\limits_0^x {xdydx} } \)

Step 4:- Integrate on y.

\(\begin{array}{l}I = \int\limits_0^1 {xy} \left| {_0^x} \right.dx.\\ = \int\limits_0^1 {{x^2}dx.} \\ = \frac{1}{3}\left( {{x^3}} \right)_0^1 = \frac{1}{3}\end{array}\)

Hence, values is \(\frac{1}{3}\)

The graph clarifies that x runs from \(y = x\) to \(x = 1\)

So, \(I = \int\limits_0^1 {\int\limits_y^1 {xdxdy} } \)

- Integrate on x

\(\begin{array}{l}I = \int\limits_0^1 {\int\limits_y^1 {xdxdy} } \\ = \int\limits_0^1 {\frac{1}{2}{x^2}\left| {_y^1dy = \frac{1}{2}\int\limits_0^1 {1 - {y^2}dy} } \right.} \end{array}\)

Step 6:- Integrate on y

\(\begin{array}{l}I = \frac{1}{2}\left( {y - \frac{1}{3}{y^3}} \right)_0^1\\ = \frac{1}{2}\left( {1 - \frac{1}{3} - 0} \right)\\ = \frac{1}{3}\\\end{array}\)

Therefore, answer is \(\frac{1}{3}\)