91Ó°ÊÓ

We will use the method of disks. Since we are revolving about the x-axis, we will integrate with respect to x. Our disk's area when we revolve it around the x-axis will be represented by the formula: \(A(x) = \pi y^2\), where y is given by the curve equation. 1. Replace \(y\) in the area with the curve equation: \(A(x) = \pi \left(\frac{1}{\sqrt{4-x^2}}\right)^2\) 2. Simplify the area formula: \(A(x) = \pi \left(\frac{1}{4-x^2}\right)\) 3. Now, we can integrate the area with respect to x to find the volume: \(V = \int_{a}^{b} A(x) \, dx\) 4. Our limits of integration are from \(x = 0\) to \(x = \frac{3}{2}\): \(V = \int_{0}^{\frac{3}{2}} \pi \left(\frac{1}{4-x^2}\right) \, dx\) 5. We can now compute the integral using the arctan formula for integration (\(\int \frac{1}{a^2 + x^2}dx = \frac{1}{a}\arctan\frac{x}{a}\) ): \(V = \pi \left[\frac{1}{2}\arctan\frac{x}{2}\right]_0^{\frac{3}{2}}\) 6. Evaluate the integral: \(V = \pi \left(\frac{1}{2}\arctan\frac{3}{4}\right)\) So the volume of the solid generated by revolving the region about the x-axis is \(\pi \left(\frac{1}{2}\arctan\frac{3}{4}\right)\).

This time we will use the method of washers. Since we are revolving about the y-axis, we need to integrate with respect to y. We also need to find the curve equation in terms of y, i.e., x as a function of y. The washer's volume when we revolve it around the y-axis will be represented by the formula: \(A(y) = \pi\left(R^2 - r^2\right)\), where R is the outer radius and r is the inner radius. 1. From the curve equation, \(y = \frac{1}{\sqrt{4-x^2}}\), solve for x: \(x = \sqrt{4 - \frac{1}{y^2}}\) 2. Determine the radii of the washer. The outer radius (R) is when x is from 0 to 3/2, and the inner radius (r) is from 0 to 3/2. \(R = \sqrt{4 - \frac{1}{y^2}}\), and \(r = 0\) 3. Replace \(R\) and \(r\) in the area formula: \(A(y) = \pi\left(\left(\sqrt{4 - \frac{1}{y^2}}\right)^2 - 0^2\right)\) 4. Simplify the area formula: \(A(y) = \pi (4 - \frac{1}{y^2})\) 5. Now, we can integrate the area with respect to y to find the volume: \(V = \int_{a}^{b} A(y) \, dy\) 6. To find the limits of integration, replace x = 0 and x = 3/2 in the curve equation: \(y(0) = 1\), and \(y\left(\frac{3}{2}\right) = \frac{2}{\sqrt{7}}\)  7. Compute the integral: \(V = \int_{\frac{2}{\sqrt{7}}}^{1} \pi(4 - \frac{1}{y^2}) \, dy\) 8. Integrate: \(V = \pi \left[4y+\frac{1}{y}\right]_{\frac{2}{\sqrt{7}}}^{1}\) 9. Evaluate the integral: \(V = \pi \left(4 - \frac{7}{2}\right)\) So the volume of the solid generated by revolving the region about the y-axis is \(\pi \left(\frac{1}{2}\right)\).