Practice Problems for Module 3
Covers Sections 6.3 and 6.4
Exercises
Directions: You should try to solve each problem first, and then click "Reveal Answer" to check your answer. You can click "Watch Video" if you need help with a problem.
1. Set up an integral representing the volume of the solid obtained by rotating the region bounded by the following curves about the line \(x=5\): \(y=3x−x^2\) and \(y=3x−9.\)
An integral giving the volume is \( \displaystyle \int_{-3}^3 \! 2\pi \left(9-x^2\right)(5-x)\, dx\)
2. Set up an integral representing the volume of the solid obtained by rotating the region bounded by the following curves about the line \(y = 5\): \(y =\sqrt{x−4}\), the \(x\)-axis and \(x = 8.\)
An integral giving the volume is \( \displaystyle \int_{0}^2 \! 2\pi (5-y)\left(4-y^2\right)\, dy\)
3. Suppose a spring has natural length of 3 ft and it takes 10 ft-lb to stretch it from 5 ft to 8 ft.
- How much work is required to stretch the spring from 4 ft to 7 ft?
- How far beyond its natural length would a force of 3 lb keep the spring stretched?
- 150/21 ft-lb
- 63/20 ft
4. A hemispherical tank has a radius of 10 m with a 2 m spout at the top of the tank. The tank is filled with water to a depth of 7 m. Set up an integral that would compute the work required to pump all the water out of the spout. Use the fact that the weight density of water is 9800 N/m\(^3.\)
The total work is given by \(\displaystyle \int_3^{10} 9800\pi \left( 100-y^2\right) (y+2) \, dy.\)
5. A trough with isosceles triangles as its ends is filled with water to a depth of 2 ft. The tank is 4 ft tall, 6 ft across at the top, 10 ft long and has a 3 ft spout. Set up an integral that would compute the work required to pump all the water out of the spout. Use the fact that the weight density of water is 62.5 lb/ft\(^3\).
Video Errata: When labeling the picture on the left (the one in 3D), the speaker said to start the axes at the top of the water. This is not correct: start it at the top of the tank as he did in the right-hand drawing, looking at the tank from the side. Also, the speaker used the wrong weight density: since we are working in ft, it needs to be 62.5 ft/lb\(^3,\) NOT 9800 N/m\(^3.\)
The total work is given by \(\displaystyle \int_2^{4} 62.5 \left[ 15(4-y)\right](y+2) \, dy.\)