Question
Use both the Shell and Disk Methods to calculate the volume of the solid obtained by rotating the region under the graph of f(x) = 5 - x^3 for 0 lessthanorequalto x lessthanorequalto 5^1/3 about the x > axU and the y-axis. Using the disk method. the volume D_s of the solid obtained by rotating the region about the x-axis is integral^a_0 g(x) dx (this is the initial integral when you setup the problem), where a = g(x) = D_1 = Using the shell method. the volume S_s of the solid obtained by rotating the region about the x-axis is integral^b_0 h(y)dy (this is the initial integral when you setup the problem), where b = h(y) = S_1 = Using the disk method. the volume 0, of the solid obtained by notating the region about the y-axis is integral^A_0 G(y)dy (this is the initial integral when you setup the problem), where A = G(y) = D_1 = Using the shell method, the volume S_y of the solid obtained by rotating the region about the y-axis is integral^B_0 H(x)dx (this is the initial integral when you setup the problem), where B = H(x) = S_y = Answer(s) submitted: (incorrect) A water storage tank has the shape of a cylinder with diameter 10 ft. It is mounted so that the circular cross-sections are vertical. If the depth of the water is 7 ft. what percentage of the total capacity is being used? Percentage % Answer(s) submitted: (incorrect) Approximate the arc length of the curve y = 1/4 x^4 over the interval [1, 2] using the trapezoidal rule and 10 intervals (i.e.. T_10). T_10 = Answer(s) submitted: (incorrect) Find the area of the surface obtained by rotating the curve y = x^3/6 + 1/2x, 1/2 lessthanorequalto x lessthanorequalto 1, about the x-axis. Area = Answer(s) submitted: (incorrect) Use Simpson's Rule w ith n = 10 to approximate the area of the surface obtained by rotating the curve y = ln(x). 1 lessthanorequalto x lessthanorequalto 3, about the x-axis. Include at least live decimal places in your answer. Area = Answer(s) submitted: (incorrect) A swimming pool is built in the shape of a rectangular parallelepiped 12 ft deep, 20 ft w ide and 20 ft long. (a) if the pool is filled to 1 ft below the top, how much work is required 10 pump all the water into a drain at the top edge of Ihe pool? (Use 62.4 lb/ft^2 for the weight density of water.) Work = ft middot lb. (b) A one-horsepower motor can do 550 ft middot lb of work per second. What size motor is required to empty the pool in 1 hour? size of motor = hp. Answer(s) submitted: (incorrect) Assume that 30 ft-lb of work is required to stretch a spring 4 ft beyond its natural length. What is the spring constant? k = lb/ft Answer(s) submitted: (incorrect)
Explanation / Answer
Solution:(33)
The capacity is proportional to the circular cross section area of the cylinder.
The percentage of the total capacity is comparing the area of the circle with water to the whole area r²,
which is = * 52 = 25.
x2 + y2 = 25 with diameter 10.
The diameter is 10, area under the curve, semi-circle is from -5 to 5.
If the depth of the water is 7 ft,
Let's find the area under the curve from y = -5 to y= 2.
We integrate the area between the function f(y) and x = 0
After we found the area, we times 2 to get the whole area.
f(y) = x = (25 - y2) from [-5,2]
A = (25 - y2) dy [-5,2]
A = (52 - y2) dy [-5,2]
Let y = 5 sin() => dy = 5 cos() d
A = (52 - 52 sin2()) 5 cos() d
= 25 cos() d
= (25/2) {1 + cos(2)} d
= (25/2) { + (1/2)sin(2)}
= (25/2) { + sin() cos()}
= (25/2) {sin-1(y/5) + (y/5)(1 - (y/5)2)}
A = (25/2) sin-1(y/5) + (1/2) y (25 - y2) from [-5,2]
Thus the crossectional area is;
=> 2 (25 - y2) dy [-5,2] = 2 [(25/2) sin-1(y/5) + (1/2) y (25 - y2)]-52 = [25 sin-1(y/5) + y (25 - y2)]-52
= 25 sin-1(2/5) + 2 (25 - 22) - 25 sin-1(-5/5) + (-5) (25 - (-5)2)
= 25 sin-1(2/5) + 221 - 25 sin-1(-1)
= 25 sin-1(2/5) + 221 - 25 (-/2) = 58.723
The total volume of water in the tank is thus about 58.723 ft^3 . To see what percentage of the total capacity this is, we take
= 58.723 / (25) = 0.7477 = 74.77 %
The tank is filled to about 74.77% capacity.
