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