Question

4 find the volume of the solid which is generated by rotating the region bounded above by $y = \sqrt{x + 2}$, below by the x-axis and on the right by the line $x=2$ about the line $x=2$. $y=0$ definite integral(s):

Explanation:

Step1: Identify the method

We use the disk/washer method (or cylindrical shells, but here disk method with respect to y is easier) since we are rotating about the vertical line \( x = 2 \). First, express \( x \) in terms of \( y \). Given \( y=\sqrt{x + 2} \), we square both sides: \( y^{2}=x + 2 \), so \( x=y^{2}-2 \).

Step2: Determine the limits of integration

The region is bounded below by the \( x \)-axis (\( y = 0 \)) and we need to find where \( x = 2 \) intersects \( y=\sqrt{x + 2} \). Substitute \( x = 2 \) into \( y=\sqrt{x + 2} \), we get \( y=\sqrt{2 + 2}=\sqrt{4}=2 \). So \( y \) ranges from \( 0 \) to \( 2 \).

Step3: Set up the integral for volume

When rotating about \( x = 2 \), the radius \( r \) of each disk (in the y - direction) is the horizontal distance from \( x=y^{2}-2 \) to \( x = 2 \), which is \( r=2-(y^{2}-2)=4 - y^{2} \). The formula for the volume of a solid of revolution using the disk method (integrating with respect to y) is \( V=\pi\int_{a}^{b}[r(y)]^{2}dy \). Here, \( a = 0 \), \( b = 2 \), and \( r(y)=4 - y^{2} \). So the volume \( V=\pi\int_{0}^{2}(4 - y^{2})^{2}dy \), and the definite integral for the volume (before multiplying by \( \pi \) if we consider the area of the disk first) or just the integral for the area of the disks (to get volume we multiply by \( \pi \)) is \( \int_{0}^{2}(4 - y^{2})^{2}dy \) (and then multiply by \( \pi \) for volume). Alternatively, if we use the shell method (integrating with respect to x), the height of the shell is \( y=\sqrt{x + 2} \), the radius of the shell is \( 2 - x \), and the limits for x: when \( y = 0 \), \( 0=\sqrt{x + 2}\Rightarrow x=-2 \), and \( x \) goes from \( -2 \) to \( 2 \). The formula for shell method is \( V = 2\pi\int_{a}^{b}(\text{radius})(\text{height})dx \), so \( V=2\pi\int_{-2}^{2}(2 - x)\sqrt{x + 2}dx \). But the disk method with respect to y is more straightforward here. Let's confirm the disk method integral:

Expanding \( (4 - y^{2})^{2}=16-8y^{2}+y^{4} \), so \( \int_{0}^{2}(16 - 8y^{2}+y^{4})dy \) (and then multiply by \( \pi \) for volume).

Answer:

If we use the disk method (integrating with respect to \( y \)): The definite integral for the volume (when using \( V=\pi\int_{a}^{b}[r(y)]^{2}dy \)) is \( \pi\int_{0}^{2}(4 - y^{2})^{2}dy \), and the integral inside (the area of the disk) is \( \int_{0}^{2}(4 - y^{2})^{2}dy=\int_{0}^{2}(16 - 8y^{2}+y^{4})dy \). If we use the shell method (integrating with respect to \( x \)): \( 2\pi\int_{-2}^{2}(2 - x)\sqrt{x + 2}dx \). The most straightforward one (disk method with y) gives the definite integral for the volume as \( \pi\int_{0}^{2}(4 - y^{2})^{2}dy \) (or the integral for the area part \( \int_{0}^{2}(4 - y^{2})^{2}dy \) and then multiply by \( \pi \) for volume).