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Math Help - Calculus Trouble

  1. #1
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    Calculus Trouble

    I am currently having trouble with these calculus practice problems. I would appreciate any help in solving these. Thank you.

    1. Find the volume of a solid whose base is enclosed by the circle x^2 + y^2=1, and whose cross sections taken perpendicular to the base are equilateral triangles with bases in the xy-plane.

    2. Let R be the region enclosed by the graphs of y=ln(x^2 + 1) and y=cos(x). Setup, using integrals, but do not evaluate the following:
    a. The area of R.
    b. The base of a solid is region R. Each cross section of the solid perpendicular to the x-axis is an isosceles right triangle. Express the volume of the solid.
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  2. #2
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    Quote Originally Posted by babolat29 View Post
    I am currently having trouble with these calculus practice problems. I would appreciate any help in solving these. Thank you.

    1. Find the volume of a solid whose base is enclosed by the circle x^2 + y^2=1, and whose cross sections taken perpendicular to the base are equilateral triangles with bases in the xy-plane.
    The volume V of a solid between a and b is the integral of its cross-sectional area A(x) from a to b, or \int_{a}^{b} A(x) dx
    The area A of an equilateral triangle is \frac{\sqrt{3}s^2}{4} where s is the length of a side. Since the trianglular cross-sections have their base in a circle with radius 1, the side of the triangular cross-section at any x is twice the distance from the x-axis to a semi-circle of radius 1, or s(x) = 2\sqrt{1 - x^2}. Therefore A(x) = \frac{\sqrt{3}}{4}(2\sqrt{1 - x^2})^2 = \frac{\sqrt{3}}{4}4(1-x^2) = \sqrt{3}(1-x^2).
    Since your solid is enclosed in a circle with radius 1, you're going to integrate the area from -1 to 1, or \int_{-1}^{1} A(x) dx = \int_{-1}^{1}\sqrt{3}(1-x^2)dx which you should know how to do.
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  3. #3
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