Question

Set up iterated integrals for both order of integration

D bounded by S.

Short answer

\(\begin{aligned}{l}I = \int\limits_0^4 {y{e^{{x^2}}} - y} dy\\ = \left( {\frac{1}{2}{e^{{x^2}}} - \frac{1}{2}{y^2}\left| {_0^4} \right.} \right)\\ = \frac{1}{2}\left( {\left( {e - 16} \right) - (1 - 0)} \right) = \frac{1}{2}\left( {{e^{16}} - 17} \right)\end{aligned}\)

Step-by-step solution

Step 1:- It is easier to set up double integral after graphing the region

Observing that y runs from line \(y = x\)up to line \(y = 4\)

\(I = \int\limits_0^4 {\int\limits_x^4 {{y^2}{e^{xy}}dydx} } \)

X ranges from 0 to y.

Therefore, integral can be set up as:

\(I = \int\limits_0^4 {\int\limits_0^y {{y^2}{e^{xy}}dxdy} } \)

\(\begin{aligned}{l} = \int\limits_0^4 {{y^2}\left( {\frac{1}{y}{e^{xy}}\left| {_0^y} \right.dy} \right)} \\ = \int\limits_0^4 {{y^2}\left( {\frac{1}{y}{e^{{y^2}}} - \frac{1}{y}} \right)} dy\\ = \int\limits_0^4 {y{e^{{y^2}}} - y} dy\end{aligned}\)

Step 2:- Integrating on y.

\(\begin{aligned}{l}I = \int\limits_0^4 {y{e^{{x^2}}} - y} dy\\ = \left( {\frac{1}{2}{e^{{x^2}}} - \frac{1}{2}{y^2}\left| {_0^4} \right.} \right)\\ = \frac{1}{2}\left( {\left( {e - 16} \right) - (1 - 0)} \right) = \frac{1}{2}\left( {{e^{16}} - 17} \right)\end{aligned}\)