Question

suppose R is the finite region bounded by f(x)= sqrt x and f(x)= x/4. find the shaded area of the object we obtain when rotating R about the x axis, and the y axis.

Answer 1

To solve for the shaded areas of the object obtained through rotation, use washer method when rotating about the x-axis and shell method when rotating about the y-axis after solving for the intersection points.

Step-by-step explanation:

The question asks to find the area of the region when the bounded area between f(x) = √x and f(x) = x/4 is rotated about the x-axis and y-axis. To find these areas, we will be using the method of disks or washers for rotation about the x-axis and the method of cylindrical shells for rotation about the y-axis. First, we have to find the points of intersection for the curves by setting √x = x/4 and solving for x. We then integrate the Volume of the washer formed by the outer radius R(x) = √x and inner radius r(x) = x/4 from the lower limit to the upper limit of x.

To find the volume when rotated about the y-axis, we consider the cylindrical shell with the formula V = 2π∫ x(f(x) - g(x)) dx, where f(x) and g(x) are our functions. In this case, we will use the cylindrical shell method to calculate the volume of rotation. These volumes represent the shaded area of the objects obtained after the rotations about respective axes.

Answer 2

We want to find the shaded area of the region R that is bounded by the functions

`f(x)=√(x)\text{ and }f(x)=(x)/(4)`

First, we find the point of intersection, this occurs at a point where the functions are equal:

`\begin{gathered} √(x)=(x)/(4) \\ 4√(x)=x \end{gathered}`

Squaring both sides we get:

`\begin{gathered} 16x=x^2 \\ x^2-16x=0 \\ x(x-16)=0 \end{gathered}`

This says us that they intersect at the point x and when x=16. For finding the shaded area, we calculate the following integral:

`\begin{gathered} \int_0^(16)\int_{(x)/(4)}^(√(x))1dydx=\int_0^(16)√(x)-(x)/(4)dx \\ =\int_0^(16)x^(1/2)-(x)/(4)dx \\ =(2)/(3)x^{(3)/(2)}-(x^2)/(8)|_0^(16) \\ =(2)/(3)(16)^{(3)/(2)}-(16^2)/(8) \\ =(32)/(3) \end{gathered}`

Then, the shaded area is 32/3 units, or 10.6666... The rotations do not affect the area, as they are isometries.