Q. 58 - Calculus

Question

Find the signed volume between the graph of the function and the xy-plane over the specified region

$$\begin{gathered} f(x,y)=x{y}^3{e}^{x^2{y}^2} \\ Where R=\left\{\right(x,y\left)\right|-2\le x\le -1,-2\le y\le 0\} \end{gathered}$$

Step-by-Step Solution

Answer

The value of the volume is $-6-2e^4+8e^{16}$cubicunits

Step 1 : Given information

We are given a function over a region

$$\begin{gathered} f(x,y)=x{y}^3{e}^{x^2{y}^2} \\ Where R=\left\{\right(x,y\left)\right|-2\le x\le -1,-2\le y\le 0\} \end{gathered}$$

Step 2: Find the volume

The value of the function is positive in the given region

Hence the volume becomes

$$\begin{gathered} {\int }_{-2}^0{\int }_{-2}^{-1}x{y}^3{e}^{x^2{y}^2}dxdy \\ V=2{\int }_{-2}^0{{\left[y{e}^{x^2{y}^2}\right]}^{-1}}_{-2}dy \\ V=2{\int }_{-2}^0[y{e}^{y^2}-y{e}^{4y^2}]dy \\ V={{[2e^{y^2}-8e^{4y^2}]}_{-2}}^0 \\ V=[-6-2e^4+8e^{16}] \end{gathered}$$