x^2/9 + y^2/4 = 1 is the base of a solid. cross-sections perpendicular to the x-axis are isosceles right triangles with the leg as the base. find the volume of the solid
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2^2/9 + y^2/4 = 1
y^2/4 = 1 - x^2/9
y^2 = 4(1 - x^2/9)
The area of a cross-sectional slice is (1/2)(2y)^2 = 2y^2 = 2(4)(1 - x^2/9)
dV = 8(1 - x^2/9) dx
The ellipse stretches from x = -3 to x = 3, but we can use symmetry to make it easier to calculate the volume.
(1/2) V = 8 ∫ (1 - x^2/9) dx {from x = 0 to x = 3} = 8 (x - x^3/27)
V = 16 (x - x^3/27) ] {0, 3} = 16 [(3 - 3^3/27) - (0 - 0^3/3)] = 16 (2 - 0) = 32
y^2/4 = 1 - x^2/9
y^2 = 4(1 - x^2/9)
The area of a cross-sectional slice is (1/2)(2y)^2 = 2y^2 = 2(4)(1 - x^2/9)
dV = 8(1 - x^2/9) dx
The ellipse stretches from x = -3 to x = 3, but we can use symmetry to make it easier to calculate the volume.
(1/2) V = 8 ∫ (1 - x^2/9) dx {from x = 0 to x = 3} = 8 (x - x^3/27)
V = 16 (x - x^3/27) ] {0, 3} = 16 [(3 - 3^3/27) - (0 - 0^3/3)] = 16 (2 - 0) = 32