Yay! Reiny, you read the description of f(x) correctly.
Recall the problem posted earlier:
Let f(x)= abs of negative abs of x + pi/2 close abs and g(x)= abs of cosx.
the wording might be confusing so here is another description of the equation (the "vertical line" that I refer to is the line you draw as you draw the absolute value sign...for example "line x line" means that x is enclosed by two vertical lines...abs of x....
So here is the description of f(x):
line, negative sign, line, x, line, plus sign, pi/2, line
a)Find the area of f(x) and g(x).
b)Find the volum of solid obtained by spinning the shape from part a) around the x- axis. when x= 3pi/2?
I need help...
1. finding the area between f(x) and g(x), on the intervals [-3pi/2, 3pi/2]
Here is my work:
A=1/2 bh + integral sign
[-pi/2,pi/2] cos x dx
I got (pi^2/4)+2.
Is that correct?
2. Next, I need help finding the volume of the solid obtained by spinning the shape from problem #1 around the x-axis.
Here is my work:
V= pi integral sign [-pi/2,pi/2] (-x+pi/2)^2 - (cos x)^2 dx
When I revolve the solid around the x-axis, I noticed that there are two radii :
outer radius ---> f(x)^2
inner radius ---> g(x)^2
Area= pi (f(x)^2 - g(x)^2)
Is this the correct set up? I was a but confused how the absolute value function (f(x)) can be squared??
Also, I need help finding the volume of the solid obtained by spinning the shape around the line x=3pi/2.
Thank you!
you got quite a bit of symmetry hanging around here.
look at area between the linear and the cosine between 0 and π/2, there are equal regions on the left of the y-axis as the right side.
the linear equation is y = -x + π/2 and the cosine curve is y = cosx
so that area is
∫((-x+π/2) - cosx) dx from 0 to π/2
= [ -(1/2)x^2 + (π/2)x - sinx ] from 0 to π/2
= ....
just realized looking at my diagram that we have the same integral expression from π/2 to 3π/2 so let's just do the integral from 0 to 3π/2
= -(1/2)(3π/2)^2 + (π/2)(3π/2) - sin(3π/2) - (0+0 - sin0)
= 9π^2/8 + 3π^2/4 - (-1)
= 15π^2/8 + 1
of course we have to double that, so the area as asked for is
15π^2/4 + 2
(better check my arithmetic, I should have written it on paper first)
for the volume you would have
2π∫(-x+π/2)^2 - (cosx)^2 dx from 0 to 3π/2
I will leave it up to you to hack your way through this,
the hard part might be to integrate (cosx)^2
one version is (1/2)(x + sinxcosx)
For this part:
"for the volume you would have
2π∫(-x+π/2)^2 - (cosx)^2 dx from 0 to 3π/2 "
Do you mean the volume when revolves around the x -axis?
If so, this would produce a washer, right?
If it is a washer then is the formula..
V= pi (and not 2pi?) integral sign from 0 to 3pi/2 ...
Is the 2pi in 2π∫(-x+π/2)^2 - (cosx)^2 dx from 0 to 3π/2
Supposed to be replaced by pi??
The 2pi would be a cylindrical shell method??
I did 2π to include the part to the left of the y-axis,
because of the symmetry I mentioned, the two parts are equal, so I doubled it.
Yes it is the washer method
To find the area between two functions, in this case f(x) and g(x), on the interval [-3π/2, 3π/2], you need to follow these steps:
1. Identify the intersection points of f(x) and g(x) on the given interval. In this case, we need to find where the two functions are equal to each other.
From the description of f(x), we have:
f(x) = | -|x| + π/2 |
From the description of g(x), we have:
g(x) = | cos(x) |
To find the intersection points, we equate f(x) and g(x) and solve for x:
| -|x| + π/2 | = | cos(x) |
To simplify this equation, we consider two cases:
Case 1: -|x| + π/2 = cos(x)
Case 2: -|x| + π/2 = -cos(x)
You can now solve each case individually to find the intersection points.
2. Once you have the intersection points, you can determine the area between the two functions by integrating the absolute difference of the two functions on the interval where f(x) is greater than g(x).
The integral should have the form:
A = ∫[a, b] |f(x) - g(x)| dx
In this case, you have correctly set up the integral:
A = 1/2 ∫[-π/2, π/2] | -|x| + π/2 - cos(x) | dx
Now, you need to evaluate this integral.
For the second part of your question, to find the volume of the solid obtained by spinning the shape around the x-axis, you can use the method of cylindrical shells.
The volume of a cylindrical shell is given by the formula:
V = ∫[a, b] 2πx * h(x) dx
where x represents the distance from the axis of rotation, and h(x) represents the height of the shell at x.
In this case, the height of the shell h(x) can be found by subtracting the inner radius (g(x)) from the outer radius (f(x)). Therefore, we have:
V = ∫[a, b] π( f(x)^2 - g(x)^2 ) dx
In your case, you correctly set up the integral:
V = π ∫[-π/2, π/2] (f(x)^2 - g(x)^2) dx
You mentioned being confused about how to square the absolute value function (f(x)). When squaring f(x), you square the entire expression, including any constants or variables inside the absolute value function.
To find the volume of the solid obtained by spinning the shape around the line x = 3π/2, you would follow a similar approach. However, instead of using the x-axis as the axis of rotation, you need to use the given line x = 3π/2.
The volume of the solid would then be given by the formula:
V = ∫[a, b] 2π | x - (3π/2) | * h(x) dx
where h(x) represents the height of the shell at x, given by the difference between the outer radius (f(x)) and the inner radius (g(x)).
I hope this explanation helps clarify the steps to find the area and volume between the functions, and how to set up the integrals correctly.