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Thread: Volume of solid given base and cross sections - trouble understanding question

  1. #1
    Newbie DragonLord's Avatar
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    Volume of solid given base and cross sections - trouble understanding question

    The following is an exercise from Rogawski's Calculus: Early Transcendentals, section 6.2:

    In Exercises 9-14, find the volume of the solid with given base and cross sections.

    9.
    The base is the unit circle $\displaystyle x^2 + y^2 = 1$ and the cross sections perpendicular to the $\displaystyle x$-axis are triangles whose height and base are equal.


    I don't clearly understand the question. What exactly are the cross sections ("perpendicular" is ambiguous), and what do I need to integrate? (The answer key says $\displaystyle V =\frac{8}{3}$.)

    --DragonLord
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    For $\displaystyle x\in[-1,1]$ the base and the height of the triangle are $\displaystyle b=h=2\sqrt{1-x^2}$ so, the area is $\displaystyle A(x)=2(1-x^2)$ . Then, $\displaystyle V=\int_{-1}^{1}A(x)\;dx=\ldots=8/3$


    Fernando Revilla
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  3. #3
    Newbie DragonLord's Avatar
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    Quote Originally Posted by FernandoRevilla View Post
    For $\displaystyle x\in[-1,1]$ the base and the height of the triangle are $\displaystyle b=h=2\sqrt{1-x^2}$ so, the area is $\displaystyle A(x)=2(1-x^2)$ . Then, $\displaystyle V=\int_{-1}^{1}A(x)\;dx=\ldots=8/3$


    Fernando Revilla
    Can you give more details on how $\displaystyle b=h=2\sqrt{1-x^2}$ is derived? If possible, post a diagram so I'll understand better. Thanks,

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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by DragonLord View Post
    Can you give more details on how $\displaystyle b=h=2\sqrt{1-x^2}$ is derived?
    For every $\displaystyle x\in [-1,1]$ the correponding ordinates on $\displaystyle x^2+y^2=1$ are $\displaystyle y=\pm \sqrt{1-x^2}$ so, the base of the triangle is $\displaystyle b=\sqrt{1-x^2}-(-\sqrt{1-x^2})=2\sqrt{1-x^2}$ .


    Fernando Revilla
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  5. #5
    Newbie DragonLord's Avatar
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    Thank you!

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