Exam 7: Applications of the Integral

Let V be the volume of the solid that lies between planes perpendicular to the x-axis from x = -3 to x = 3. The cross-sections of this solid perpendicular to the x-axis run from $y = - \sqrt { 9 - x ^ { 2 } }$ to $y = \sqrt { 9 - x ^ { 2 } }$ and they are semicircles with diameters in the xy-plane. Then V is

Let A denote the area enclosed by the equations $y = x ^ { 2 }$ and $y = 2 - | x |$ Then A is

Four particles having masses 2, 3, 3, and 4 kg are located on the xy-plane at $( - 1 , - 2 ) , ( 1,3 ) , ( 0,5 ) , ( 2,1 ),$ respectively. The y-coordinate of the center of mass of this system is located at $\bar { y } =$ ?

Let A denote the area enclosed by the y-axis and the equations y = sin x and y = cos x for $0 \leq x \leq \frac { \pi } { 4 }$ Then A is

Let V be the volume of the solid that lies between planes perpendicular to the x-axis from x = 0 to x = 4. The cross-sections of this solid perpendicular to the x-axis run from $y = \frac { x ^ { 2 } } { 8 }$ to $y = \sqrt { x }$ and they are squares with bases in the xy-plane. Then V is

The arc length of $y = ( x + 1 ) ^ { \frac { 3 } { 2 } } + 1 \text { for } x \in [ 3,8 ]$ is

Let V be the volume of the solid generated by revolving the region enclosed by y = -x², y = x² - 2x; about the y-axis. Then V is

Let V be the volume of the solid generated by revolving the region enclosed byabout the x-axis. Then V is

A spring in equilibrium is 0.8 meters long and an external force of 2 Newtons stretches the spring to a length of 1.2 meters. The work done by the spring force in stretching it from equilibrium to 3 meters, in Newton-meters, is ​

A cylindrical sewer pipe of radius 3 feet is half full of water of density 62.5 pounds per cubic foot. A gate used to seal off the sewer is placed perpendicular to the pipe opening. The force, in pounds, due to hydrostatic pressure on one end of the gate is

Let V be the volume of the solid generated by revolving the region enclosed by $x = y ^ { 2 } + 1 , x = 2 y + 1;$ about $x = - 1$ Then V is

Which of these integrals can be used to calculate the surface area of the solid formed by revolving the region bounded by $f ( x ) = \sqrt [ 3 ] { x },$ the -axis, x = 1, and x = 5 about the x-axis?

Let $\bar { x }$ denote the x-coordinate of the centroid of the region enclosed by $y = 1 - x ^ { 2 }$ and $y = 0$ Then $\bar { x }$ =

Let V be the volume of the solid generated by revolving the region enclosed by about x = 1. Then V is

Let V be the volume of the solid that lies between planes perpendicular to the x-axis from x = -1 to x = 1. The cross-sections of this solid perpendicular to the x-axis run from $y = - 4 \sqrt { 1 - x ^ { 2 } }$ to $y = 4 \sqrt { 1 - x ^ { 2 } }$ and they are squares with bases in the xy-plane. Then V is

A 25-meter long, 15-meter wide, and 5-meter deep rectangular water tank is full. Assume the density of water is 1000 kg/cu m. The work required to pump all the water over the top of the tank, in Newton-meters, is

Let A denote the area enclosed by the equations $y = 2 | x |$ and $y = 3 - x$ Then A is

The arc length of $y = - \frac { 2 } { 3 } ( x + 1 ) ^ { \frac { 3 } { 2 } } \text { for } x \in [ 1,4 ]$ is

Let A denote the area enclosed by the equations $y ^ { 2 } = x - 2$ and $x = 4$ Then A is

A right circular cylindrical water tank with height 30 meters and radius 5 meters is $\frac { 2 } { 3 }$ full. Assume the density of water is 1000 kg/cu m. The work required to pump all the water over the top of the tank, in Newton-meters, is