Practice Problems for Module 2
Covers Section 6.2
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. The base of a solid is the region enclosed by \(x = y^2 − 9\) and \(x = 7\). Its cross-sections are perpendicular to the \(x\)-axis and are equilateral triangles. Set up an integral to find the volume of the solid.
An integral giving the volume is \[ \displaystyle \int_{-9}^7 \! \sqrt{3}(x+9) \, dx\]
2. The base of a solid is the region enclosed by \(y = \sqrt{x + 5}\), \(x = 4\) and the \(x\)-axis. Its cross-sections are perpendicular to the \(y\)-axis and are semicircles. Set up an integral to find the volume of the solid.
An integral giving the volume is \[\displaystyle \int_{0}^3 \! \frac{1}{8} \pi \left( 9 - y^ 2\right)^2 \, dy\]
3. Set up an integral representing the volume of the solid obtained by rotating the region bounded by the following curves about the \(x\)-axis: \(y =\dfrac{2}{x}\), the \(x\)-axis, \(x = 1\) and \(x = 4.\)
An integral giving the volume is \[\displaystyle \int_{1}^4 \! \dfrac{4\pi}{x^2} \, dx\]
4. Set up an integral representing the volume of the solid obtained by rotating the region bounded by the following curves about the line \(x=7\): \(x=y^2 +3\) and \(x=7.\)
An integral giving the volume is \[ \displaystyle \int_{-2}^2 \! \pi \left(4-y^2\right)^2\, dy\]
5. Set up an integral representing the volume of the solid obtained by rotating the region bounded by the following curves about the \(y\)-axis: \(y = \ln(x)\), the \(x\)-axis and \(x = e.\)
An integral giving the volume is \[ \displaystyle \int_{0}^1 \! \pi \left(e^2-e^{2y}\right)\, dy\]
6. Set up an integral representing the volume of the solid obtained by rotating the region bounded by the following curves about the line \(y=5\): \(y = \sqrt{x+1}\), the \(x\)-axis and \(x = 8.\)
An integral giving the volume is \[\displaystyle \int_{-1}^8 \! \pi \left[25-\left(5-\sqrt{x+1}\right)^2\right]\, dx\]