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)
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!
3 answers
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??
"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
because of the symmetry I mentioned, the two parts are equal, so I doubled it.
Yes it is the washer method