1. (12 points) Given the area $A$ in the figure,
(a) Express the area $A$ as an integral. DO NOT EVALUATE the integral.
(b) Express the volume created by rotating this area about the y-axis. DO NOT EVALUATE the integral.
Answer
(b) The volume created by rotating the area A about the y-axis can be expressed as an integral by summing up the volumes of the infinitesimally small disks that make up the volume. The thickness of each disk is dx and the radius is the function f(x). Therefore, the volume V can be expressed as the integral of \(\pi f(x)^2 dx\) from a to b, where a and b are the x-coordinates of the left and right boundaries of the area. In this case, the function f(x) is a piecewise function: it is equal to 5 for 0 <= x <= 1, and it is equal to 2 for 1 < x <= 4. Therefore, the volume V can be expressed as the integral from 0 to 1 of \(\pi (5)^2 dx\) plus the integral from 1 to 4 of \(\pi (2)^2 dx\), which in latex is \(\int_{0}^{1} \pi (5)^2 dx + \int_{1}^{4} \pi (2)^2 dx\).
Step 1 ：(a) The area A can be expressed as an integral by summing up the areas of the infinitesimally small rectangles that make up the area. The width of each rectangle is dx and the height is the function f(x). Therefore, the area A can be expressed as the integral of f(x) dx from a to b, where a and b are the x-coordinates of the left and right boundaries of the area. In this case, the function f(x) is a piecewise function: it is equal to 5 for 0 <= x <= 1, and it is equal to 2 for 1 < x <= 4. Therefore, the area A can be expressed as the integral from 0 to 1 of 5 dx plus the integral from 1 to 4 of 2 dx, which in latex is \(\int_{0}^{1} 5 dx + \int_{1}^{4} 2 dx\).
Step 2 ：(b) The volume created by rotating the area A about the y-axis can be expressed as an integral by summing up the volumes of the infinitesimally small disks that make up the volume. The thickness of each disk is dx and the radius is the function f(x). Therefore, the volume V can be expressed as the integral of \(\pi f(x)^2 dx\) from a to b, where a and b are the x-coordinates of the left and right boundaries of the area. In this case, the function f(x) is a piecewise function: it is equal to 5 for 0 <= x <= 1, and it is equal to 2 for 1 < x <= 4. Therefore, the volume V can be expressed as the integral from 0 to 1 of \(\pi (5)^2 dx\) plus the integral from 1 to 4 of \(\pi (2)^2 dx\), which in latex is \(\int_{0}^{1} \pi (5)^2 dx + \int_{1}^{4} \pi (2)^2 dx\).
