Consider the region bounded by

x2 + y2 = 64 with x ≥ 0, and y ≥ 0

. A solid is created so that the given region is its base and cross-sections perpendicular to the x-axis are squares. Set up a Riemann sum and then a definite integral needed to find the volume of the solid.

Question

Browning

Mathematics

11 October, 19:24

Consider the region bounded by

x2 + y2 = 64 with x ≥ 0, and y ≥ 0

. A solid is created so that the given region is its base and cross-sections perpendicular to the x-axis are squares. Set up a Riemann sum and then a definite integral needed to find the volume of the solid.

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Answers (1)

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Kamden · Answer 1

The equation is a circle centered at the origin with radius 8 (sqrt (64))

Therefore, the bounded region is just a quarter circle in the first quadrant.

Riemann Sum: ∑⁸ₓ₋₋₀ (y²) Δx=∑⁸ₓ₋₋₀ (64-x²) Δx

Definite Integral: ∫₀⁸ (y²) dx=∫₀⁸ (64-x²) dx

Consider the region bounded byx2 + y2 = 64 with x ≥ 0, and y ≥ 0. A solid is created so that the given region is its base and cross-sections perpendicular to the x-axis are squares. Set up a Riemann sum and then a definite integral needed to find the volume of the solid.

Consider the region bounded by

x2 + y2 = 64 with x ≥ 0, and y ≥ 0

. A solid is created so that the given region is its base and cross-sections perpendicular to the x-axis are squares. Set up a Riemann sum and then a definite integral needed to find the volume of the solid.