The quantity of the strong is the indispensable under $latex z= f( x, y)$, with the wall surfaces provided by the aircrafts $latex y= 0$, $latex z= 0$ and also $latex x= 3$.

The primary step is to figure out the crossways of $latex f( x, y)= 4-y ^ 2$ with the coordinate axes, in order to locate the limitations of assimilation.

** Junction of $latex f( x, y)$ with z-axis**

Taking $latex y = 0$, we have: $latex z =4-0 ^ 2= 4$

Therefore, the junction of the allegorical cyndrical tube with the z-axis is the factor $latex (0,0,4)$.

** Junction of $latex f( x, y)$ with y-axis**

Taking $latex z= 0$, we have:

$ latex 4-y ^ 2= 0Rightarrow y =2$

The favorable origin is taken considering that the junction with the favorable y-axis is looked for, which is the factor $latex (0,2,0)$.

The aircraft $latex y= 0$ restricts the strong, considering that it remains in the very first octant, according to the declaration, consequently, the base of the strong is a rectangular shape with measurements $latex x= 3$ and also $latex y = 2$, while both continuing to be wall surfaces are provided by the aircrafts $latex x= 0$ and also $latex x= 3$.

The resulting strong is received the number listed below:

The quantity of the strong represents the indispensable:

$$ V= int_a ^ bint_c ^ d f( x, y) dydx$$

Where $latex y$ ranges $latex 0$ and also $latex 2$, while $latex x$ ranges $latex 0$ and also $latex 3$, consequently:

$$ V= int_0 ^ 3left[int_0^2 (4 -y^2)dyright]dx$$

First, the internal indispensable is determined:

$$ int_0 ^ 2 (4 -y ^ 2) dy= int_0 ^ 2 4dy-int_0 ^ 2 y ^ 2dy$$

$$ 4yBig|_ 0 ^ 2-dfrac {y ^ 3} {3} Huge|_ 0 ^ 2$$

$$= 8-dfrac {8} {3} =dfrac {16} {3} $$

The outcome is replaced in the quantity indispensable brace:

$$ V= int_0 ^ 3left( dfrac {16} {3} right) dx$$

$$= dfrac {16} {3} int_0 ^ 3dx= left( dfrac {16} {3} right) xBig|_ 0 ^ 3= 16$$