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Math Help - Help on a confusing Integration question

  1. #1
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    Help on a confusing Integration question

    I've got one question left on my assignment, and I'm not too sure what the questions even asking.

    Here's the question:


    Can somebody help me understand this question? And is the question asking about 2D or 3D models? I can understand how 2D might be possible to work out, but have no idea about 3D, since the height (z-axis) is not specified.

    Maybe I'm just looking at the question the wrong way! Any help is accepted .
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  2. #2
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    Quote Originally Posted by Phatmat View Post
    I've got one question left on my assignment, and I'm not too sure what the questions even asking.

    Here's the question:


    Can somebody help me understand this question? And is the question asking about 2D or 3D models? I can understand how 2D might be possible to work out, but have no idea about 3D, since the height (z-axis) is not specified.

    Maybe I'm just looking at the question the wrong way! Any help is accepted .
    a): for a fixed 0 \leq x \leq \pi find the area of an equilateral triangle with each side of length 6 \sqrt{\sin x}. then integrate the result on the interval [0,\pi] to find the volume. it's as simple as that!

    b): the same idea as a).
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  3. #3
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    I'm still confused..

    What do you mean by:
    "then integrate the result on the interval to find the volume. it's as simple as that!"?

    Is the triangle only 2D then? Because how can each side of the triangle be of length 6sqrt(sin x) when that's a semi-circle shape?
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  4. #4
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    Quote Originally Posted by Phatmat View Post
    I'm still confused..

    What do you mean by:
    "then integrate the result on the interval to find the volume. it's as simple as that!"?

    Is the triangle only 2D then? Because how can each side of the triangle be of length 6sqrt(sin x) when that's a semi-circle shape?
    for a fixed x_0, the triangle is in the plane x=x_0. you leaned in high school or earlier that the area of an equilateral triangle with each side of length a is \frac{\sqrt{3}}{4}a^2.

    so in your example the area of each cross section is A(x)=\frac{\sqrt{3}}{4}(6 \sqrt{\sin x})^2=9\sqrt{3} \sin x. the volume you're looking for is \int_0^{\pi} A(x) \ dx= 18 \sqrt{3}.
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  5. #5
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    Thanks for your help, your answer was right. I used the same method for the square (part b), and it was right aswell.
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