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let R be the region bounded by the graphs of y= sin(pi)(x) and y=x^3-4c, as shown in the figure below. A) Find the volume of the solid formed when the region R is revolved around the line y=2. B) the region is the base of a solid. For this solid each cross section perpendicular to the x-axis is an isosceles right triangle whose hypotenuse is the base. Find the volume of this solid.

let R be the region bounded by the graphs of y sinpix and yx34c as shown in the figure below A Find the volume of the solid formed when the region R is revolved class=


Answer :

Answer:

A) V = ∫₀² π [(2 − x³ + 4x)² − (2 − sin πx)²] dx ≈ 77.762

B) V = ∫₀² ¼ (sin πx − x³ + 4x)² dx ≈ 2.495

Step-by-step explanation:

A region rotated about an external axis results in a hollow shape, the volume of which can be found using washer method. To use this method, we slice the volume into thin washers. By writing the inside radius and outside radius in terms of y₁ and y₂, and the thickness in terms of x, we can write an expression for the volume of the washer, then write an integral to find the total volume of the solid.

For part B, we do something similar, again slicing the volume into cross sections. This time, the cross sections are triangles instead of circles. Still, the volume of each cross section is equal to the area times the thickness. By integrating, we add up the individual cross sections to get the total volume.

A) Revolving the region about y = 2 yields a hollow donut-ish shape. By cutting a thin slice perpendicular to this axis, we get a washer with an inner radius of r and an outer radius of R. Writing in terms of y₁ and y₂:

r = 2 − y₁

R = 2 − y₂

The thickness of the washer is dx. The volume of the washer is therefore:

dV = π (R² − r²) dx

dV = π [(2 − y₂)² − (2 − y₁)²] dx

dV = π [(2 − x³ + 4x)² − (2 − sin πx)²] dx

Integrating from x=0 to x=2, the total volume of the revolute is:

V = ∫ dV

V = ∫₀² π [(2 − x³ + 4x)² − (2 − sin πx)²] dx

Evaluating with a calculator, this volume is approximately 77.762.

B) This time, instead of rotating the region about an axis, the region is the base of a solid with cross sections of right isosceles triangles, where the hypotenuse of the triangle is the base. The length of this hypotenuse is:

s = y₁ − y₂

For a 45-45-90 triangle, the sides are equal to the hypotenuse divided by √2. Therefore, the cross sectional area is:

A = ½ bh

A = ½ (s/√2) (s/√2)

A = ¼ s²

A = ¼ (y₁ − y₂)²

For a thin slice of thickness dx, the volume is:

dV = A dx

dV = ¼ (y₁ − y₂)² dx

So the total volume is:

V = ∫ dV

V = ∫₀² ¼ (y₁ − y₂)² dx

V = ∫₀² ¼ (sin πx − x³ + 4x)² dx

Evaluating with a calculator, this volume is approximately 2.495.

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