Determining the Volume of a Solid of Revolution
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Key Questions

The volume of a solid of revolution for a curve
#y = f(x)# rotated around the#x# axis can be found through:#V = pi int_a^b y^2 dx#
To solve this integral you must find the bounds#a# and#b# .For example, the curve
#y=x^2 + 2# is rotated between the bounds#x = 1# and#x = 2# is rotated about the#x# axis. Find the volume of the solid formed.
Since#y = x^2 + 2#
#y^2 = x^4 + 4x^2 + 4#
#:. V= pi int_1^2 x^4 + 4x^2 + 4 dx#
#= pi [ (x^5)/(5) + (4x^3)/(3) +4x]_1^2#
#= (431pi)/5# or#270.81# (2dp) units cubedWhen the curve is rotated around the
#y# axis the volume of the solid is given by:#V = pi int_e^d x^2 dx#
The function must be in the form where#x# is a function of#y# and the bounds#d# and#e# must be read off the#y# axis. 
The short and sweet version of the answer is:
#(pi)*int_0^1((4x^2)^2(4x)^2)dx# =#1.2pi# or approximately#3.769911 . . .# This is the region bounded by the points
#(0,0)# and#(1,1)# since those points lie on both the graph of the line (#g(x)=x# ) and the parabola (#(f(x)=x^2# ). We will take this region and subdivide it into vertical regions that are one can picture as tiny rectangle. When the rectangles are spun about the line#y=4# , they will form tiny washers. If there are#n# subintervals, then the width of each subinterval is#Delta x=1/n# . The outer radius (distance from the function farther away from the line of rotation to the line of rotation) is#(4f(x))# . The inner radius (distance from the function nearer to the line of rotation to the line of rotation) is #(4g(x))The usual setup for washers is:
#*lim_(nrarroo)sum_(i=1)^n((piR(x))^2(pir(x))^2)*(Deltax)# which becomes#(pi)*int_a^b((piR(x)^2(pir(x))^2)*(dx)# which for our functions equals:#int_0^1(pi(4f(x))^2pi(4g(x))^2)dx# =#(pi)*int_0^1((4x^2)^2(4x)^2)dx# =#1.2pi# or approximately#3.769911 . . .# 
A cone with base radius
#r# and height#h# can be obtained by rotating the region under the line#y=r/hx# about the xaxis from#x=0# to#x=h# .
By Disk Method,
#V=pi int_0^h(r/hx)^2 dx={pi r^2}/{h^2}int_0^hx^2 dx#
by Power Rule,
#={pir^2}/h^2[x^3/3]_0^h={pir^2}/{h^2}cdot h^3/3=1/3pir^2h# 
If the radius of its circular cross section is
#r# , and the radius of the circle traced by the center of the cross sections is#R# , then the volume of the torus is#V=2pi^2r^2R# .Let's say the torus is obtained by rotating the circular region
#x^2+(yR)^2=r^2# about the#x# axis. Notice that this circular region is the region between the curves:#y=sqrt{r^2x^2}+R# and#y=sqrt{r^2x^2}+R# .By Washer Method, the volume of the solid of revolution can be expressed as:
#V=pi int_{r}^r[(sqrt{r^2x^2}+R)^2(sqrt{r^2x^2}+R)^2]dx# ,
which simplifies to:
#V=4piR\int_{r}^r sqrt{r^2x^2}dx#
Since the integral above is equivalent to the area of a semicircle with radius r, we have
#V=4piRcdot1/2pi r^2=2pi^2r^2R#
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Applications of Definite Integrals

1Solving Separable Differential Equations

2Slope Fields

3Exponential Growth and Decay Models

4Logistic Growth Models

5Net Change: Motion on a Line

6Determining the Surface Area of a Solid of Revolution

7Determining the Length of a Curve

8Determining the Volume of a Solid of Revolution

9Determining Work and Fluid Force

10The Average Value of a Function