Let R be the first quadrant region enclosed by the graph of y= 2e^-x and the line x=k.
a) Find the area of R in terms of k.
b) Find the volume of the solid generated when R is rotated about the x-axis in terms of k.
c) What is the volume in part (b) as k approaches infinity?
To find the area of the region R in terms of k:
Step 1: Set up the integral limits for the region R.
Since R is the first quadrant region enclosed by the graph of y = 2e^-x and the line x = k, we need to find the intersection point of the two curves.
Setting y = 2e^-x equal to the equation of the line x = k, we have:
2e^-x = k
Taking the natural logarithm of both sides gives:
-ln(2) - x = ln(k)
Simplifying, we find:
x = -ln(k) - ln(2)
So the integral limits for the region R are x = 0 (the y-axis) and x = -ln(k) - ln(2) (the intersection point).
Step 2: Set up the integral expression for the area of R.
The area of R can be calculated using the integral of the function y = 2e^-x from the lower limit to the upper limit:
Area = ∫[0, -ln(k) - ln(2)] 2e^-x dx
Step 3: Evaluate the integral.
Integrating 2e^-x gives:
Area = [2(-e^-x)] evaluated from 0 to -ln(k) - ln(2)
= 2(-e^-(-ln(k) - ln(2))) - 2(-e^0)
= 2(-e^(ln(k) + ln(2))) - 2(-1)
= 2(-e^(ln(k)) * e^(ln(2))) + 2
= 2(-k * 2) + 2
= -4k + 2
So the area of R in terms of k is -4k + 2.
To find the volume of the solid generated when R is rotated about the x-axis in terms of k:
Step 1: Set up the integral limits for the volume.
Since R is being rotated around the x-axis, the volume can be calculated using the disk method. The radius of the disk is given by the function, y = 2e^-x.
The integral limits for the volume are the same as for the area calculation, x = 0 and x = -ln(k) - ln(2).
Step 2: Set up the integral expression for the volume.
The volume V can be calculated using the integral of the function for the area of a circle (πr^2) from the lower limit to the upper limit:
V = ∫[0, -ln(k) - ln(2)] π(2e^-x)^2 dx
= ∫[0, -ln(k) - ln(2)] π(4e^-2x) dx
= π∫[0, -ln(k) - ln(2)] 4e^-2x dx
Step 3: Evaluate the integral.
Integrating 4e^-2x gives:
V = π[(-1/2)e^-2x] evaluated from 0 to -ln(k) - ln(2)
= π[(-1/2)e^-2(-ln(k) - ln(2))) - (-1/2)e^-2(0)]
= π[(-1/2)e^(2ln(k) + 2ln(2))] - π(-1/2)
Expanding the exponential term, we get:
V = π[(-1/2)(e^(ln(k)^2) * e^(ln(2)^2))] - π(-1/2)
= π[(-1/2)(k^2 * 2^2)] - π(-1/2)
= π[(-1/2)(4k^2)] - π(-1/2)
= -2πk^2 + π/2
So the volume of the solid generated when R is rotated about the x-axis in terms of k is -2πk^2 + π/2.
To find the volume as k approaches infinity:
As k approaches infinity, the term -2πk^2 dominates, so the volume approaches negative infinity.
Therefore, the volume as k approaches infinity is negative infinity.
To find the area and volume in terms of k, we first need to locate the points of intersection between the graph of y = 2e^-x and the line x = k.
To find the point of intersection, we set the equations equal to each other and solve for x:
2e^-x = k
Now, let's solve each part of the question step by step:
a) Finding the Area of R in terms of k:
To find the area of R, we need to calculate the integral of 2e^-x from x = 0 to x = k. This will give us the area enclosed by the graph of y = 2e^-x and the line x = k:
Area_R = ∫[0,k] 2e^-x dx
To evaluate this integral, we can use integration by substitution.
Let u = -x, du = -dx.
Now, when x = 0, u = 0, and when x = k, u = -k.
Substituting the values and shifting the limits of integration:
Area_R = ∫[-k,0] 2e^u (-du)
= -∫[0,-k] 2e^u du
= -2∫[0,-k] e^u du
= -2[e^u]_[0,-k]
= -2[e^-k - e^0]
= -2[e^-k - 1]
= 2(1 - e^-k)
Therefore, the area of R in terms of k is given by 2(1 - e^-k).
b) Finding the Volume of the Solid when R is Rotated about the x-axis in terms of k:
To find the volume of the solid, we need to calculate the integral of the cross-sectional area of R from x = 0 to x = k, as the graph is rotated around the x-axis:
Volume = ∫[0,k] π(2e^-x)^2 dx
= ∫[0,k] 4πe^(-2x) dx
To evaluate this integral, we can use integration by substitution.
Let u = -2x, du = -2dx.
Now, when x = 0, u = 0, and when x = k, u = -2k.
Substituting the values and shifting the limits of integration:
Volume = ∫[0,-2k] 4πe^u (-du/2)
= -2π∫[0,-2k] e^u du
= -2π[e^u]_[0,-2k]
= -2π[e^(-2k) - e^0]
= -2π[e^(-2k) - 1]
Therefore, the volume of the solid when R is rotated about the x-axis in terms of k is given by -2π[e^(-2k) - 1].
c) Finding the Volume as k approaches infinity:
As k approaches infinity, the exponential term e^(-2k) approaches zero because the exponent becomes infinitely negative. Therefore, the volume can be approximated as:
Volume ≈ -2π(-1) = 2π
Therefore, as k approaches infinity, the volume of the solid approaches 2π.
a) For the first quadrant region, x>0 to x = k, and the enclosed area is
Integral y dx =
Integral 2e^-x dx
x = 0 to k
= -2 e^-k + 2 e^0
= 2(1 - e^-k)
b) Make the integrand pi*y^2 dx and perform the resulting integration from 0 to k
c) This should be obvious after doing (b)