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

    To see the full video page and find related videos, click the following link.
    WIR 20B M152 V11


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

    To see the full video page and find related videos, click the following link.
    WIR 20B M152 V12


  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 \( \displaystyle \int_{1}^4 \! \dfrac{4\pi}{x^2} \, dx\). 

    To see the full video page and find related videos, click the following link.
    WIR 20B M152 V13


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

    To see the full video page and find related videos, click the following link.
    WIR 20B M152 V14


  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. 
     

    To see the full video page and find related videos, click the following link.
    WIR 20B M152 V15


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

    To see the full video page and find related videos, click the following link.
    WIR 20B M152 V16