BC Calc Ch. 7

2/13/2016

Post comments here relevant to Chapter 7 only.

Jeffrey W

2/19/2016 09:25:13 pm

7.3 #15
We know that if you integrate the area of a shape''s cross section(s) over a specific length you can obtain that shape's volume. For this situation, we will use the equation we are given and it's distance fro the y-axis (Which we are revolving around) as the radius of a circular cross section.

Givens:
Area of a circle=πr^2
Equation (In terms of y)= tan(π/4y)=r

Integral set up:
Int A(x)
Int πr^2
π Int (tan(π/4y))^2 dy from 0->1

Answer:
0.858 units^3

Jeffrey W

2/19/2016 09:50:06 pm

Displacement is the numerical value given by the position function when you plug in a point in time. This means that at that moment, whatever object you are keeping track of is the output's distance away from your reference point. Displacement has a negative and positive direction

Total distance is how far the object has traveled. It has no direction and is just a summation of the absolute value of it's displacement over a certain length of time.

Example:
Let's say a particle moves to the right 5 units, but then moves to the left 2 units. It's displacement would be 3 units because that is hoe far the particle is from it's initial starting position, but it's total distance would be 7, because the object traveled 5 units, then another 2.

Jeffrey W

2/19/2016 09:50:51 pm

*how

Kavya Anjur

2/22/2016 09:34:59 pm

As learned in 7.1, the strategy to midweek with integrals is to first approximate what you want to find as a Reimann sum of values of a continuous function multiplied by interval lengths. With f(x) as the function and subintervals of delta x along the interval [a, b], the sums will be the approximation of a standard reimann sum. Next, a definite integral should be written of f(x) from a to b to express the limit of these sums as the norms of the partitions go to zero. The last step is to solve the integral numerically or with antiderivatives.

Kavya Anjur

2/22/2016 09:36:28 pm

**strategy to model with integrals

Kavya Anjur

2/23/2016 10:43:33 pm

For 7.2, problem #6, you can the region's symmetry to solve the problem. You would find the area of the shade region analytically by taking the area between the curves, so x^2 - (-2x^4), which is equal to x^2 + 2x^4. You would do two times the integral of (x^2 + 2x^4) from 0 to 1 by using the region's symmetry. If you then solve the integral by taking the antiderivative, you would get 2[(1/3)x^3 + (2/5)x^5] from 0 to 1, which is equivalent to 22/15.

Ananth Putcha

2/24/2016 06:53:16 pm

When you are finding the area of a solid with a washer as it's cross section, you take the integral of π(big radius^2) - π(small radius^2)

Ananth Putcha

3/17/2016 11:23:19 pm

Example:
Regiom bounded by y-axis, Y=X and X=3 revolved around y=-1 from
π × integral from 0 to 3 of ((x^2)^2)-(1)^2

Brandon C

2/26/2016 09:22:55 pm

7.2 #3
int(y^3-y^2)dy from 0 to 1
F(x) = (y^4)/4 - (y^3)/3 evaluated from 0 to 1
F(b)-F(a) = (1/4)-(1/3) - 0 = (-1/12) u^2

Qiyuan Zhou link

3/2/2016 05:45:54 pm

p.392 7.3 #13

Find the volume of the solid generated by revolving x+2y=2 {0≤x≤2} around the x axis.

First, we must be aware of the cross sectional shape of the solid we are will generate. This function will not result in any "holes" in the solid generated from revolving it around the given line, so the area of one of its cross sections at point x can be found with the formula π((-x/2)+1)^2

Integrating the cross sectional area across the domain that we are revolving the function will give us volume of the solid.

Therefore, we integrate π((-x/2)+1)^2dx over the interval [0,2].

After performing the integral, we are left with the task of evaluating (π/12)x^3(-πx/2)+x from 0 to 2.

This results in a volume of 2/3 π units^3

Ananth Putcha

3/7/2016 10:16:32 pm

7.1 #7 v(t)=(e^sint)cost, 0≤t≤2π

a. Stopped at π/2, 3π/2
Moving left from π/2 to 3π/2
Moving right from 0 to π/2 and 3π/2 to 2π

Rachel Willy

3/8/2016 04:13:49 pm

Work = Force * Distance

Hooke's law formula is used to determine the force constant, displacement and force in a stretched spring.

Force = kx

x is the displacement in the spring in meters
k is the spring constant or force constant

Paige Eber

3/8/2016 09:29:17 pm

Definition; Area Between Curves

If f(x) and g(x) are continuous with f(x)>g(x) or f(x)=g(x) throughout [a,b] hen the area between the curves y=f(x) and y=g(x) from a to b is the integral of [f(x) - g(x)] from a to b

to find the actual set up mathmatically and not written out go to page 375 in your book! sorry i cant post it here

Ananth Putcha

3/8/2016 09:59:53 pm

7.2 #18 Find the area enclosed by x=y^2 and x=y+2.

Step 1: make y^2=y+2 in order to find where they intersect.
(y^2)-y-2=0; y=-1,2.

Step 2: Take the integral of the top function - the bottom function with respect to y using the points of intersection as the bounds of the integral.
Integral from -1 to 2 of (y+2) - (y^2)dy = 4.5

Answer: 4.5 units^2

Christopher Glenn

3/8/2016 11:22:32 pm

7.5 #8
Given: natural length of 10in 800lbs of force to stretch spring to 14in
a) calculate the force constant
800=k*4
k=200 lb*in
b) find work done in stretching spring from 10 to 12
[int]from 0 to 2 of 200x so we evaluate 100x^2 from 0 to 2
W = 400 lb * in
c) find how far beyond natural length it is stretched if 1600lbs of force stretch it.
1600=200x
x=8in

Taylor Garcia

3/9/2016 09:10:27 pm

7.2 #14
find the area of the regions enclosed by the lines and curves
y = (x^4) - 4(x^2) + 4
y = x^2

find where they intersect by setting them equal to each other
(x^4) - 4(x^2) + 4 = x^2
(x^4) - 3(x^2) + 4 = 0
factor:
x = -2, -1, 1, 2

since the graphs shows that the function from 0 to 2 is identical from -2 to 0, we can just take the integral from 0 to 2 multiplied by 2
also remember to take the top curve minus the bottom curve
since this graph is split into 2 sections, we take the area of one plus the area of the other

((∫ from 0 to 1 ((x^4) - 4(x^2) + 4) - x^2 dx) + (∫ from 1 to 2 x^2 - ((x^4) - 4(x^2) + 4) dx)) * 2

(2.533 + 1.466) * 2
8 units squared

Taylor Garcia

3/10/2016 06:25:49 pm

7.3 #17
find the volume of the solid generated by rotating about the x axis
y = x^2
y= 0
x = 2

if you graph all of these, you will notice that the graph created by the function x^2 is "on top" of the line created by x = 2.

They intersect at x = 2
π∫ from 0 to 2 (x^2)^2 dx
π ((x^5)/5) from 0 to 2
π ((32/5) - (0))
6.4π units cubed

Taylor Garcia

3/14/2016 09:06:31 pm

7.3 #23
find the volume of the solid about the x axis

y = (x^2) + 1
y = x + 3

x + 3 = (x^2) + 1
(x^2) - x - 2 = 0
(x-2)(x+1)
x = -1, 2

π∫ from -1 to 2 of ((x+3)^2) - (((x^2)+1)^2) dx
73.513 units cubed

Elise T

3/14/2016 11:57:12 pm

To find the volume of a solid:
the volume of a solid of known integrable cross section area A(x) from x=a to x=b is the integral of A from a to b, V=int(a to b) of A(x)dx

Shreyas Mohan

3/15/2016 11:25:13 am

Cross Sections:
In my mind, cross sections are pretty much the same as rotating a graph around the axis, except they may not be circular. So pretty much, instead of finding the "radius" of a circle, you may have to find the "height" of the triangular cross section or the "side length" of the square cross section. Once you have modeled and equation for how this length changes across the axis, you can integrate the same way you would if you were dealing with a rotated graph. However, you won't be using (pi)(r^2) in the integral, you'll have to use the appropriate area equation for the shape of the cross section.

Taylor Garcia

3/15/2016 09:36:29 pm

7.3 #27
find the volume of the solid rotated about y = 2^(1/2)
y = 2^(1/2)
y = sec(x)tan(x)

sec(x)tan(x) = 2^(1/2)
x = 0.785

π∫ from 0 to 0.785 of ((sec(x)tan(x) - (2^(1/2)))^2) - ((2^(1/2) - 2^(1/2))^2) dx
2.301 units cubed

Taylor Garcia

3/16/2016 10:17:43 pm

7.3 #29
Find the volume of the solid rotated about the y axis
x = (5^(1/2))y^2
x = 0
y = -1
y = 1

2π∫ from 0 to 1 of ((5^(1/2))y^2)^2 dy
6.263 units cubed

Kevin Knox

3/17/2016 11:05:54 pm

7.3 #20

Find the volume of the solid generated by revolving the region bounded by the lines and curves about the x-axis.

y = x - x^2
y = 0

y = x - x^2 reaches y = 0 at x = 0 and x = 1

The area of the cross section is equal to πr^2
r = x - x^2
A = π (x - x^2)^2
A = π (x^2 - 2x^3 + x^4)

V = ∫(0-->1) A dx
V= ∫(0-->1) π (x^2 - 2x^3 + x^4) dx
V = π [ (1/3)x^3 - (1/2)x^4 + (1/5)x^5 ](0 --> 1]
V = π ( 1/3(1)^3 - 1/2(1)^4 + 1/5(1)^5 ) - π (0)
V = π ( 1/3 - 1/2 +1/5 )
V = π ( 10/30 - 15/30 + 6/30 )
V = π /30

Michelle H

3/18/2016 11:30:21 pm

Integrals can find:
Work
Fluid pressure
Probability

Allison Y

3/21/2016 05:27:13 pm

7.2 #11
find the area of the region enclosed by y = x^2 – 2 and y = 2
x^2 – 2 = 2
x = ±2 , so the curves intersect at x=2 and x= -2
∫[2 – (x^2 – 2)]dx from -2 to 2
= ∫(4 – x^2)dx from -2 to 2
take antiderivative:
= [4x – (x^3 / 3)] from 2 to -2
F(2) – F(-2)
= (8 – (8/3)) – (-8 + (8/3))
= 32/3 u^2

Allison Y

3/22/2016 07:21:22 pm

7.2 #8
find the area of the region enclosed by y = x^2 and x+y = 2
integrate with respect to y:
∫[(2 – y) - √y ] dy from 0 to 1
take antiderivative:
= [2y – (y^2 / 2) – (2y^(3/2) / 3)] from 0 to 1
F(1) – F(0)
= (2 - (1/2) – (2/3)) – 0
= 5/6 u^2

Paige Eber

3/24/2016 04:51:55 pm

Occasionally we are able to use geometry to integrate, like with the function y=x. If we wanted to integrate this function over the interval [0,2] we would first recognize that the area we need to find is a right triangle with a base of 2 and a height of 2. So we can just use the formula for an area of a triangle and find the area that way

A = (1/2)bh
A = (1/2)(2)(2)
A = 2

so the integral from 0 to 2 of x = 2

Elise T

3/25/2016 12:51:28 pm

7.2 #5
Find the area between the curves from -2 to 2 when the functions are y=2x^2 and y=(x^4)-(2x^2)

int(-2→2) of (2x^2)-((x^4)-(2x^2))
antiderivative is ((2/3)x^3)-((x^5)/5 - (2/3)x^3)
subtract the function with x=-2 from x=2, and you get (128/15)

Paige Eber

3/25/2016 05:57:27 pm

7.2 #3
Find the area between x = y^3 and x = y^2 from 0 to 1

the integral from 0 to 1 of [y^2 - y^3]
the anti derivative is [y^3/3 - y^4/4]
1/3-1/4 - [0/3-0/4]
1/3-1/4
4/12-3/12

1/12 is your answer

Shreyas Mohan

3/26/2016 08:47:03 am

Something interesting to think about is how energy and work are interrelated and how calculus can be used to prove physics topics. Most of us have learned in physics class that elastic energy is 1/2kx^2 and that F=kx. Since work is the integral of force with respect to time. W=1/2kx^2 as well, which is the same equation as energy. This shows that work is the same as a change in energy.

Kevin Knox

3/27/2016 10:02:48 pm

7.2 #1

Find the area of the shaded region. Unfortunately, I cannot show you the graph. However, I can describe it.

The given plane has the graph y = cos^2 (x) and the graph y = 1. The shaded region extends from x = 0 to x = π. It fills the area between y = 1 and y = cos^2 (x)

∫(0 --> π) ( 1 - cos^2(x) )dx
cos^2 (x) = 1/2(cos(2x)) + 1/2
∫(0 --> π) ( 1 - (1/2(cos(2x)) + 1/2) )
∫(0 --> π) ( 1/2 - 1/2(cos(2x)) )
[ 1/2(x) - 1/4(sin(2x)) ] (0 --> ∫(0 --> π))
( 1/2(π) - 1/4(0) ) - ( 1/2(0) - 1/4(0) )
π/2

∫(0 --> π) ( 1 - cos^2(x) )dx = π/2

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Mrs. Johnson's 2015-2016 BC Calculus Center for Review.

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BC Calc Ch. 7

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