# Finding the Area and the Volume of portions under curves.

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Finding the Area and the Volume of portions under curves.

 Normally, to find the area under a single curve, you could use Riemann's sums. However our way is faster and less aggravating.  There are 3 steps to finding the area under a single curve. 1)Find your intervals 2)Take an integral 3)Solve using your intervals

 Y=x 2 [0,1]x-axis

 X=y 2 [0,1]y-axis

 To find the area under 2 curves, the steps are similar as the ones used to find the area under just a single curve.  In the method, you have to establish which variable you are using.  If the functions you are using are in terms of X then you subtract the function on the bottom from the function on top.

 Your intervals are the numbers at the top and bottom of the integral sign. They are the numbers that you are limiting your function to.  This is helpful for finding the area because you need to have a finite portion of the graph that you are trying to find.

 To find your intervals, you must simply set the 2 of your functions equal to each other. Then you simplify them so you have your intervals.

 y=x 2 y= [0,1]

 Y=3xy=x 3 +2x 2  Give the intervals and the solution  first quadrant.

 In these functions, you will be in terms of Y. Since you are in terms of Y, your intervals change from being on the x-axis to on the Y- axis.  You figure out which function is further on the left of the plain and subtract it from the one further on the right  Right – Left.

 X= y 2 -6yX=-Y[0,5]

 x=y 2 -5yy-axis  Give your intervals, and solution

 Put graph

 if you try and find the area of sine between 0- 2  then you will end up getting zero.  That’s because the area that you get that is above the x-axis, mathematically, will cancel out the other.  This is why you take half the interval and then multiply the final product by 2.  So go from 0-  and multiply your solution by 2.

 y=x 3 y=x x=1x=-1 

 y=x 2 y=-2x 4  x=1 x=-1

 y=x 2 -4y=-x 2 -2x  Give the interval, and the solution.

 x=y 2 x=y+2  Give the intervals, and the solution.

 Y=sinx[0,2  ]

 To find the volume of a solid of revolution, you will need to know four things: the function, what axis you are rotating the function around, the interval which you are finding the volume and what equation to use.

 You use the disk method when the whole interval of the function is touching the axis you are rotating it around.  The equation for the disk method is π b ∫ a [f(x)] 2 dx  The function you are finding the volume for must have its variables equaling the axis you are rotating it around.

 Also, if the axis you are rotating the function around is the x-axis, your function should be equal to y, and if you are rotating the function around the y-axis, your function must equal x.  For example, if you were rotating x 2 around the x- axis, it would appear as y=x 2 and if it was around the y-axis, you would have to solve for x.  Once you plug the function into the equation, you take its integral and solve at the interval.

 Here is an example using the disk method around the x-axis: π 0 ∫ 4 [f(x 2 )] 2 dx

 Here is an example using the disk method around the y-axis: π 0 ∫ 4 [f(√y)] 2 dx

 The washer method is used when there are two functions and there will be a hole in the revolution of the solid.  You always subtract the inside function from the outside function.  The equation for the washer method is π b ∫ a [(f(x)) 2 - (g(x)) 2 ]dx  Depending on the axis of rotation, the function will be equal to the opposite variable of the axis.

 For example, if you are rotating the volume about the y-axis, the functions would both be equal to x, and vice versa.  If you were finding the volume y=x 2 minus y=x 3 about the x-axis, you wouldn’t have to change the variables. But if you were rotating them about the y- axis, you would have to solve for x.

 Here is an example using the washer method about the x-axis: π 0 ∫ 4 [(e x ) 2 -(x) 2 ]dx

 Here is an example using the washer method about the y-axis: π 0 ∫ 1 [(y) 2 -(√y) 2 ]dx

π 0 ∫ 4 [(x 3 ) 2 -(x) 2 ] dx about x-axis

 The shell method is used when you are solving for the volume of one function is going to touch the axis of revolution at one point or when the function would be a disk method around the opposite axis.  The equation for the shell method is 2π b ∫ a [radius*height]dx  The radius of the function is how wide it is and its height is the function. Usually the radius is x or y, depending on what you set the function equal to.

 If the function was y=√x, the radius of the function is x and the height is √x.  When using the shell method, the function must be equal to the opposite axis you are revolving the function around.  For example, if your function was y=√x, you would use the shell method if you were revolving it around the x-axis, otherwise you would use the disk method.

 Here is an example of using the shell method about the y-axis: 2π 0 ∫ 1 [x(2x-x 2 )]dx

 Here is an example using the shell method about the x-axis: 2π 0 ∫ 3 [y(√y)]dx

2π 0 ∫ 4 [x(x 2 -4x)]dx about the y-axis

 http://mathdemos.gcsu.edu/mathdemos/shel lmethod/gallery/gallery.html http://mathdemos.gcsu.edu/mathdemos/shel lmethod/gallery/gallery.html  http://www.mathgv.com/ http://www.mathgv.com/  http://www.musopen.org/music.php http://www.musopen.org/music.php

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