Charles C. answered 05/18/21
Adjunct Math Professor, Calculus and Linear Algebra focus
the base is a filled circle of radius 5 in the y-z plane. let the variable of integration be y which we evaluate from -5 to 5. now we need area of the equilateral triangles relative to y. the base of the triangle is the length of the chord, c, of the circle at distance y from the center/origin. bi-sect the chord to the center to get a right triangle with sides y and 1/2 of the chord, with radius 5 as the hypotenuse. thus (1/2 c)^2 + y^2 = 25. solving for c = 2sqrt(25 - y^2). the area of an equilateral triangle of side c is A = [sqrt(3)/4] c^2. so we sub for c with it's expression in y to get the area of triangle = [sqrt(3)(25 - y^2)]. now we evaluate the integral [sqrt(3)(25 - y^2)] dy from -5 to 5 = 500sqrt(3)/3.