Question

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

$$\begin{gathered} f(x,y)=-2x^2{y}^3 \\ WhereR=\left\{\right(x,y\left)\right|-2\le x\le 3,-1\le y\le 5\} \end{gathered}$$

Short answer

The value of the integral is$1686.66cubicunits$

Step-by-step solution

Given information

We are given a rectangular region

$$\begin{gathered} f(x,y)=-2x^2{y}^3 \\ WhereR=\left\{\right(x,y\left)\right|-2\le x\le 3,-1\le y\le 5\} \end{gathered}$$

Find the volume

We have,

$$\begin{gathered} f(x,y)>0when{R}_1=\left\{\right(x,y\left)\right|-2\le x\le 0,-1\le y\le 0\left\} \\ And \\ f\right(x,y)<0when{R}_2=\{(x,y)|0\le x\le 3,0\le y\le 5\} \end{gathered}$$

Also we have,

$$\begin{gathered} {\int }_R-2x^2{y}^{3}dA={\int }_{R_1}-2x^2{y}^{3}dA-{\int }_{R_2}-2x^2{y}^{3}dA \\ ={\int }_{-1}^0{\int }_{-2}^0-2x^2{y}^{3}dxdy-{\int }_0^3{\int }_0^5-2x^2{y}^{3}dxdy \\ =\left[\frac{-2x^3{y}^4}{12}\right]-\left[\frac{-2x^3{y}^4}{12}\right] \\ =-\frac{16}{12}+\left[\frac{20250}{12}\right] \\ =1686.66cubicunits \end{gathered}$$