Consider the region between the graph of $f ( x ) = 1 - x ^ { 2 }$ and the line y = 2 on the interval [0, 1].
-Use disk/washer method to construct definite integrals to find the volume of the solid that is obtained by revolving this region around the vertical line x = 2.
A) 2.38 $\pi$
B) 2.83 $\pi$
C) 3.83 $\pi$
D) 4.83 $\pi$

Question

Quizplus · Accepted Answer

The volume of the solid is found by integrating the area of washers formed by revolving the region around x = 2. The outer radius is $2 - x$ and the inner radius is $2 - x$ for $y = 2$ and $2 - \sqrt{1 - x^2}$ for $f(x)$. The integral is $\pi \int_0^1 [(2-x)^2 - (2-\sqrt{1-x^2})^2] \, dx$, which evaluates to 3.83$\pi$.

Consider the Region Between the Graph Of f(x)=1−x2f ( x ) = 1 - x ^ { 2 }f(x)=1−x2

Consider the Region Between the Graph Of $f ( x ) = 1 - x ^ { 2 }$