Question

In Exercises $55-58$, find the signed volume between the graph of the given function and the $xy$-plane over the specified rectangle in the $xy$-plane.

localid="1651244649922" role="math" $f\left(x,y\right)=x^2-y^3+4$,

where localid="1651244677121" $R=\left\{\left(x,y\right)|0\le x\le 3and-2\le y\le 3\right\}$.

Short answer

The required volume is $\frac{225}{4}$.

Step-by-step solution

Step 1. Given Information

We have given the following function :-

$f\left(x,y\right)=x^2-y^3+4$

where $R=\left\{\left(x,y\right)|0\le x\le 3and-2\le y\le 3\right\}$

We have to find the volume between the graph of the function $f\left(x,y\right)$and the specified rectangle in the $xy-$plane.

Step 2. To find Volume

We have given the following function :-

$f\left(x,y\right)=x^2-y^3+4$,

where localid="1650628129086" $R=\left\{\left(x,y\right)|0\le x\le 3and-2\le y\le 3\right\}$

Then the signed volume will be as following :-

$\int {\int }_R\left|x^2-y^3+4\right|dA={\int }_{-2}^3{\int }_0^3\left(x^2-y^3+4\right)dxdy$

Then by using Fubini's theorem, we have :-

${\int }_{-2}^3{\int }_0^3\left(x^2-y^3+4\right)dxdy={\int }_{-2}^3\left({\int }_0^3\left(x^2-y^3+4\right)dx\right)dy$

Then :-

localid="1651245884124" $$\begin{gathered} {\int }_{-2}^3\left({\int }_0^3\left(x^2-y^3+4\right)dx\right)dy \\ ={\int }_{-2}^3{\left[\frac{x^3}{3}-x{y}^3+4x\right]}_0^{3}dy \\ ={\int }_{-2}^3\left[\left(\frac{27}{3}-3y^3+12\right)-0\right]dy \\ ={\int }_{-2}^3\left(9-3y^3+12\right)dy \\ ={\int }_{-2}^3\left(21-3y^3\right)dy \\ ={\left[21y-\frac{3}{4}{y}^4\right]}_{-2}^3 \\ =\left[\left(63-\frac{3}{4}\times 81\right)-\left(-42-\frac{3}{4}\times 16\right)\right] \\ =63-\frac{243}{4}+42+12 \\ =117-\frac{243}{4} \\ =\frac{468-243}{4} \\ =\frac{225}{4} \end{gathered}$$