Question

Find the signed volume between the graph of the given function and the xy-plane over the specified rectangle in the xy-plane

$$\begin{gathered} f(x,y)=y^3{e}^{x{y}^2} \\ WhereR=\left\{\right(x,y\left)\right|0\le x\le 2,-2\le y\le 3\} \end{gathered}$$

Short answer

The volume is$16414245cubicunits$

Step-by-step solution

Given information

We are given an integral as

$$\begin{gathered} f(x,y)=y^3{e}^{x{y}^2} \\ WhereR=\left\{\right(x,y\left)\right|0\le x\le 2,-2\le y\le 3\} \end{gathered}$$

Evaluate the integral

We have

$$\begin{gathered} f(x,y)>0when{R}_1=\left\{\right(x,y\left)\right|0\le x\le 2,0\le y\le 3\left\} \\ and \\ f\right(x,y)<0when{R}_2=\{(x,y)|0\le x\le 2,-2\le y\le 0\} \end{gathered}$$

Therefore

$$\begin{gathered} {\int }_R{y}^3{e}^{x{y}^2}dA={\int }_{R_1}{y}^3{e}^{x{y}^2}dA-{\int }_{R_2}{y}^3{e}^{x{y}^2}dA \\ ={\int }_0^3{\int }_0^2{y}^3{e}^{x{y}^2}dxdy-{\int }_{-2}^0{\int }_0^2{y}^3{e}^{x{y}^2}dxdy \\ ={\int }_0^3{{\left[y{e}^{x{y}^2}\right]}^2}_{0}dy-{\int }_{-2}^0{{\left[y{e}^{x{y}^2}\right]}^2}_{0}dy \\ ={\int }_0^3(y{e}^{2y^2}-y)dy-{\int }_{-2}^0(y{e}^{2y^2}-y)dy \\ =16414245cubicunits \end{gathered}$$