Volume of Sold with a Known Crosssection

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Discussion Overview

The discussion revolves around calculating the volume of a solid formed by isosceles right triangles that are perpendicular to the x-axis, with the base defined by the equation x^2 + y^2 = 4. Participants explore the area formula for the triangles and how it relates to the volume calculation.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant inquires about the appropriate area formula to use for the triangles, suggesting the formula for the area of a triangle as bh/2.
  • Another participant explains that the hypotenuse of the isosceles right triangle is y, leading to a derivation of the area based on the legs of the triangle, resulting in an area of y.
  • A different participant challenges the area calculation, proposing that the legs of the triangle can be expressed as a = sqrt(1/2) * y, leading to an area of (1/4)y^2.
  • The same participant reiterates their area calculation, and another participant confirms that their conclusion is correct.

Areas of Agreement / Disagreement

There is some agreement on the area calculation, but the derivation of the area formula and the interpretation of the triangle dimensions show some uncertainty and differing approaches among participants.

Contextual Notes

The discussion includes various interpretations of the triangle dimensions and area calculations, which may depend on the definitions used and assumptions made about the geometry involved.

mmg0789
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the base is bounded by x^2 + y^2 = 4
the solid is formed by isosceles right triangles perpendicular to x axis. the hypotenuse of each triangle is sitting on the base of the figure

i just need to know what area formula i would start out with, is it just bh/2 ?
 
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An "isosceles right triangle perpendicular to the x axis" has hypotenuse y. Since it is isosceles, a2+ a2= 2a2= y2. That is, each leg has length a= y\sqrt{2} and so its area is (1/2)bh= (1/2)(a*a)= (1/2)(2y)= y.
 
hmm i found that a= sqrt(1/2) * y from 2a^2=y^2, the area would then be
(1/4)y^2. Right?
 
mmg0789 said:
hmm i found that a= sqrt(1/2) * y from 2a^2=y^2, the area would then be
(1/4)y^2. Right?

yes, you are right.
 

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