 # 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.
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 $$\displaystyle \int_{-9}^7 \! \sqrt{3}(x+9) \, dx$$.

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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 $$\displaystyle \int_{0}^3 \! \frac{1}{8} \pi \left( 9 - y^ 2\right)^2 \, dy$$.

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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: $$y = 2/x$$, the $$x$$-axis, $$x = 1$$ and $$x = 4$$.

An integral giving the volume is $$\displaystyle \int_{1}^4 \! \dfrac{4\pi}{x^2} \, dx$$.

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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 $$\displaystyle \int_{-2}^2 \! \pi \left(4-y^2\right)^2\, dy$$.

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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 $$\displaystyle \int_{0}^1 \! \pi \left(e^2-e^{2y}\right)\, dy$$.
Video Errata: The presenter repeatedly says disk, but should have said washer.

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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 $$\displaystyle \int_{-1}^8 \! \pi \left[25-\left(5-\sqrt{x+1}\right)^2\right]\, dx$$.

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