1. The region enclosed by the x-axis and the parabola is revolved about the line x = –1 to generate the shape of a cake. What is the volume of the cake? DO NOW

2. Section 7.3d

3. The region enclosed by the x-axis and the parabola is revolved about the line x = –1 to generate the shape of a cake. What is the volume of the cake? x = –1 Let’s first try to integrate with respect to y  Washers!!! To find the inner and outer radii of our washer, we would need to get the equation of the original parabola in terms of x... not an easy task…

4. The region enclosed by the x-axis and the parabola is revolved about the line x = –1 to generate the shape of a cake. What is the volume of the cake? Instead of slicing horizontally, we will cut a series of cylindrical slices by cutting straight down all the way around the solid. The radii of the cylinders gradually increase, and the heights of the cylinders follow the contour of the parabola: smaller to larger, then back to smaller. When we unroll one of these shells, we have a rectangular prism, with height equal to the height of the shell, length equal to the circumference of the shell, and thickness equal to.

5. The region enclosed by the x-axis and the parabola is revolved about the line x = –1 to generate the shape of a cake. What is the volume of the cake? Volume of each shell: Here, we have To find the total volume of the solid, integrate with respect to x:

6. The region bounded by the curve, the x-axis, and the line x = 4 is revolved about the x-axis to generate a solid. Find the volume of the solid. Let’s solve this problem first by integrating with respect to x: Cross section area: Volume:

7. The region bounded by the curve, the x-axis, and the line x = 4 is revolved about the x-axis to generate a solid. Find the volume of the solid. Now let’s integrate with respect to y using our new technique: Radius of each shell: Height of each shell: Limits of integration: 0 to 2 Thickness of each shell:

8. The region bounded by the curve, the x-axis, and the line x = 4 is revolved about the x-axis to generate a solid. Find the volume of the solid. Now let’s integrate with respect to y using our new technique: Volume:

9. Let’s redo a problem from a previous class, this time using shells… Find the volume of the solid generated by revolving the region bounded above by the square root function and below by the identity function about the y-axis Cross section area: Volume: Our previous solution

10. Let’s redo a problem from a previous class, this time using shells… Find the volume of the solid generated by revolving the region bounded above by the square root function and below by the identity function about the y-axis Shell radius: Volume: Shell height: Shell thickness:

11. Let’s redo a problem from a previous class, this time using shells… Find the volume of the solid generated by revolving the triangular region bounded by the lines y = 2x, y = 0, and x = 1 about the line x = 2. R r Cross section area: Volume: Our previous solution

12. Let’s redo a problem from a previous class, this time using shells… Find the volume of the solid generated by revolving the triangular region bounded by the lines y = 2x, y = 0, and x = 1 about the line x = 2. Shell radius: Shell height: Shell thickness: Volume: