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Find the volume of the solid formed by rotating the region bounded by the given curves about the indicated axis of revolution. (Round your answer to two decimal places.) x = y2, 2y − x + 3 = 0; about the y-axis

User RMachnik
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The volume of the solid is approximately 1.94 cubic units.

Sketch the region and axis of rotation.

First, graph the curve x = y^2.

Then, rewrite the equation 2y - x + 3 = 0 as a function of y: x = 2y - 3.

Shade the region enclosed by the two curves and the y-axis.

Picture the region being rotated about the y-axis.

Identify a representative washer.

Consider a thin slice of the solid perpendicular to the y-axis.

The slice will be a washer with a hole in the middle.

The radius of the washer will be equal to the distance between the curve x = y^2 and the y-axis.

The radius of the hole will be equal to the distance between the curve x = 2y - 3 and the y-axis.

Express the radius of the washer and the hole as functions of y.

Radius of the washer: r_1(y) = y^2

Radius of the hole: r_2(y) = (2y - 3)

Set up the definite integral.

The volume of each washer is π[(r_1(y))^2 - (r_2(y))^2] dy.

To find the total volume, we integrate this expression over the interval of integration, which is the range of y-values that define the shaded region.

Evaluate the definite integral.

The lower bound is the y-value where the curves intersect, which can be found by setting the functions equal to each other and solving for y: y^2 = 2y - 3 --> y = 1 or y = -3.

We choose the positive y-value, y = 1, as the lower bound.

The upper bound is the point where the curve x = 2y - 3 intersects the y-axis, which is y = 3/2.

Therefore, the definite integral is: V = π ∫_1^(3/2) [(y^2)^2 - ((2y - 3))^2] dy

Evaluate the integral using integration by parts: V ≈ 1.94

Therefore, the volume of the solid is approximately 1.94 cubic units.

Find the volume of the solid formed by rotating the region bounded by the given curves-example-1
User Aquiseb
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