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Practice Problems for Module 2
Practice Problems for Module 2
Section 6.2
Directions.
The following are review problems. We recommend you work the problems yourself, and then click "Answer" to check your answer. If you do not understand a problem, you can click "Video" to learn how to solve it. You can also follow the link for the full video page to find related videos.
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.
Answer
An integral giving the volume is \( \displaystyle \int_{-9}^7 \! \sqrt{3}(x+9) \, dx\).
Video
To see the full video page and find related videos, click the following link.
WIR 20B M152 V11
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.
Answer
An integral giving the volume is \( \displaystyle \int_{0}^3 \! \frac{1}{8} \pi \left( 9 - y^ 2\right)^2 \, dy\).
Video
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WIR 20B M152 V12
Set up an integral representing the volume of the solid obtained by rotating the region bounded by the following curves about the \(x\)-axis: \(y = 2/x\), the \(x\)-axis, \(x = 1\) and \(x = 4\).
Answer
An integral giving the volume is \( \displaystyle \int_{1}^4 \! \dfrac{4\pi}{x^2} \, dx\).
Video
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WIR 20B M152 V13
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 \(x=7\).
Answer
An integral giving the volume is \( \displaystyle \int_{-2}^2 \! \pi \left(4-y^2\right)^2\, dy\).
Video
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WIR 20B M152 V14
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 \(x = e\).
Answer
An integral giving the volume is \( \displaystyle \int_{0}^1 \! \pi \left(e^2-e^{2y}\right)\, dy\).
Video
Video Errata:
The presenter repeatedly says disk, but should have said washer.
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WIR 20B M152 V15
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 \(x = 8\).
Answer
An integral giving the volume is \( \displaystyle \int_{-1}^8 \! \pi \left[25-\left(5-\sqrt{x+1}\right)^2\right]\, dx\).
Video
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WIR 20B M152 V16