Respuesta :
Answer:
[tex]100\pi\int\limits^9_0 {(\sqrt y)^2(14-y)} \, dy[/tex] ft-lbs.
Step-by-step explanation:
Given:
The shape of the tank is obtained by revolving [tex]y=x^2[/tex] about y axis in the interval [tex]0\leq x\leq 3[/tex].
Density of the fluid in the tank, [tex]D=100\ lbs/ft^3[/tex]
Let the initial height of the fluid be 'y' feet from the bottom.
The bottom of the tank is, [tex]y(0)=0^2=0[/tex]
Now, the height has to be raised to a height 5 feet above the top of the tank.
The height of top of the tank is obtained by plugging in [tex]x=3[/tex] in the parabolic equation . This gives,
[tex]H=3^2=9\ ft[/tex]
So, the height of top of tank is, [tex]y(3)=H=9\ ft[/tex]
Now, 5 ft above 'H' means [tex]H+5=9+5=14[/tex]
Therefore, the increase in height of the top surface of the fluid in the tank is given as:
[tex]\Delta y=(14-y)[/tex] ft
Now, area of cross section of the tank is given as:
[tex]A(y)=\pi r^2\\r\to radius\ of\ the\ cross\ section[/tex]
Radius is the distance of a point on the parabola from the y axis. This is nothing but the x-coordinate of the point.
We have, [tex]y=x^2[/tex]
So, [tex]x=\sqrt y[/tex]
Therefore, radius, [tex]r=\sqrt y[/tex]
Now, area of cross section is, [tex]A(y)=\pi (\sqrt y)^2[/tex]
Work done in pumping the contents to 5 feet above is given as:
[tex]W=D\int\limits^{y(3)}_{y(0)} {A(y)(\Delta y)} \, dy[/tex]
Plug in all the values. This gives,
[tex]W=100\int\limits^9_0 {\pi (\sqrt y)^2(14-y)} \, dy\\\\W=100\pi\int\limits^9_0 { (\sqrt y)^2(14-y)} \, dy\textrm{ ft-lbs}[/tex]
