A.Find the area of the region bounded above by y=2cosx and above by y=secx,-π/4≤x≤π/4. B.Find the volume of the sold generated by revolving the region in (A) above about the x-axis
A. usingsymmetry, the area is just
∫[0,π/4] 2cosx - secx dx
= √2 - 2tanh-1 π/8
Using discs of thickness dx for the volume,
v = 2∫[0,π/4] π(R^2-r^2) dx
where R=2cosx and r=secx
v = 2∫[0,π/4] π((2cosx)^2-(secx)^2) dx = π^2
Using shells of thickness dy, we have to split the region where the curves intersect.
v = 2∫[1,√2] 2πrh dy
where r=y and h=x=arcsec(y)
+ 2∫[√2,2] 2πrh dy
where r=y and h=x=arccos(y/2)
v = 2∫[1,√2] 2πy*arcsec(y) dy
+ 2∫[√2,2] 2πy*arccos(y/2) dy
= π^2-2π + 2π
= π^2
Hi Sammy this may sound extremely outrageous and I don't know if your active but do you know the answers for gifted and talented L.A. At connexus for unit 2 sounds and ideas unit test closed book it's 20 questions and I'm so confused
A. To find the area of the region bounded above by y=2cosx and above by y=secx, we need to find the area between the curves.
We first need to find the intersection points of the two curves:
For y = 2cosx, we have:
y = secx (1/secx) = cosx
So, 2cosx = cosx
Simplifying and solving for x:
cosx = 0
x = π/2, 3π/2
To find the area, we integrate y2 - y1 with respect to x over the given interval:
A = ∫(2cosx - secx) dx, π/4 ≤ x ≤ 3π/4
Integrating 2cosx:
A = 2sinx - ln|secx + tanx| + C
Evaluating the definite integral:
A = [2sin(π/4) - ln|sec(π/4) + tan(π/4)|] - [2sin(π/4) - ln|sec(3π/4) + tan(3π/4)|]
A = [2(√2/2) - ln|√2 + 1|] - [2(√2/2) - ln|√2 - 1|]
A = √2 - ln(√2 + 1) - √2 + ln(√2 - 1)
A = -ln(√2 + 1) + ln(√2 - 1)
A = ln(√2 - 1) - ln(√2 + 1)
Therefore, the area of the region bounded above by y=2cosx and above by y=secx, -π/4 ≤ x ≤ π/4, is ln(√2 - 1) - ln(√2 + 1).
B. To find the volume of the solid generated by revolving the region from part A about the x-axis, we'll use the disk method.
The radius of the disks will be the y-coordinate, y = 2cosx.
The volume of each disk is given by V = πr^2Δx, where Δx is the thickness of the disks.
To find the volume of the entire solid, we integrate the volume of the disks with respect to x over the given interval:
V = ∫π(2cosx)^2 dx, -π/4 ≤ x ≤ π/4
V = ∫4cos^2(x) π dx
Using the double-angle identity, cos^2(x) = (1 + cos(2x))/2:
V = ∫2(1 + cos(2x)) π dx
V = 2π ∫(1 + cos(2x)) dx
V = 2π(x + (1/2)sin(2x)) + C
Evaluating the definite integral:
V = 2π[(π/4) + (1/2)sin(2(π/4))] - 2π[(-π/4) + (1/2)sin(2(-π/4))]
V = 2π(π/4 + (1/2)sin(π/2)) - 2π(-π/4 + (1/2)sin(-π/2))
V = 2π(π/4 + (1/2)) - 2π(-π/4 + (-1/2))
V = (π/2 + π) - (-π/2 - π)
V = π + π/2 + π/2
V = 2π
Therefore, the volume of the solid generated by revolving the region from part A about the x-axis is 2π.
To find the area of the region bounded above by y = 2cos(x) and below by y = sec(x), -π/4 ≤ x ≤ π/4, we can use integration.
A. Area of the region:
1. Draw the graph of y = 2cos(x) and y = sec(x) in the given range, -π/4 ≤ x ≤ π/4.
2. Determine the points of intersection between the two graphs, which will provide the boundaries for the region.
3. Integrate the difference between the upper curve (sec(x)) and the lower curve (2cos(x)) with respect to x, from the lower bound (point of intersection) to the upper bound (point of intersection).
The integral for the area is: A = ∫[lower bound, upper bound] (sec(x) - 2cos(x)) dx
B. To find the volume of the solid generated by revolving the region in part A about the x-axis, we can use the method of cylindrical shells.
1. For each value of x in the range -π/4 ≤ x ≤ π/4, find the corresponding y-values on the curve y = 2cos(x) and y = sec(x).
2. For each x-value, calculate the circumference (2πr) of the shell by using the formula for the circumference of a circle, where r is the y-coordinate.
3. Calculate the height of each shell by subtracting the value of y from the x-axis (y=0).
4. The volume of each shell is given by multiplying the circumference by the height, giving you the volume of that specific shell.
5. Sum up the volumes of all the shells by integrating using the range -π/4 ≤ x ≤ π/4, and summing the volumes of each shell.
The integral for the volume is: V = ∫[lower bound, upper bound] (2πr * h) dx, where r = y-coordinate and h = height of the shell.
Note: Calculating the exact bounds and evaluating the integrals is best done using numerical methods or a computer algebra system.