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Chapter 6: Problem 39

Let \(R\) be the region bounded by the following curves. Use the disk or washer method to find the volume of the solid generated when \(R\) is revolved about the \(y\) -axis. $$x=\sqrt{4-y^{2}}, x=0$$

Short Answer

Expert verified
Question: Calculate the volume of the solid formed by revolving the region between the equations \(x = \sqrt{4 - y^2}\) and \(x = 0\) about the y-axis. Answer: The volume of the solid is \(32\pi\) cubic units.

Step by step solution

01

Determine the limits of integration

The limits of integration can be found by identifying the intersection points of the two given curves along the y-axis. The curves intersect when x = 0: $$0 = \sqrt{4 - y^2} \Rightarrow y^2 = 4 \Rightarrow y = -2, 2$$ Thus, the limits of integration are from \(-2\) to \(2\).
02

Set up the integral using the disk method

In this case, since we are revolving around the y-axis, we will calculate the volume of each disk-like slice as a function of y. The radius of each disk will be the horizontal distance from the y-axis (x-coordinate) and the thickness will be dy. The volume of each disk is given by the formula: $$dV = \pi (radius)^2(thickness) \Rightarrow dV = \pi x^2 dy$$ To use the disk method, it is necessary to rewrite the equation \(x = \sqrt{4 - y^2}\) as a function of y. So, we have: $$x = \sqrt{4 - y^2}$$ Now, substitute x in the volume formula: $$dV = \pi (\sqrt{4 - y^2})^2 dy$$
03

Integrate the volume formula

Now, we integrate the volume formula with respect to y from the limits of integration determined in step 1: $$V = \int_{-2}^{2} \pi (4 - y^2) dy$$
04

Evaluate the integral

Evaluate the definite integral: $$V = \pi\int_{-2}^2(4-y^2)dy = \pi \left[ 4y - \frac{1}{3}y^3 \right]_{-2}^2$$ Now, substitute the limits: $$V = \pi ( \left[4\cdot(2) - \frac{1}{3}(2)^3\right] - \left[4\cdot(-2) - \frac{1}{3}(-2)^3\right] ) = \pi(16 - (-16)) = 32\pi$$ Thus, the volume of the solid generated when the region is revolved about the y-axis is \(32\pi\) cubic units.

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