The region R is bounded by the x-axis, y-axis, x = 3 and y = 1/(sqrt(x+1))
A. Find the area of region R.
B. Find the volume of the solid formed when the region R is revolved about the x-axis.
C. The solid formed in part B is divided into two solids of equal volume by a plane perpendicular to the
x-axis. Find the x-value where this plane intersects the x-axis.
Please show your work!
A.
∫[0,3] 1/√(x+1) dx
= 2√(x+1) [0,3]
= 2
B. using discs, v=∫πy^2 dx
= π∫[0,3] 1/(x+1) dx
= πlog(x+1) [0,3]
= πlog4
C. We want c where
π∫[0,c] 1/(x+1) dx = π∫[c,3] 1/(x+1) dx
log(c+1)-log(0+1) = log(3+1)-log(c+1)
2log(c+1) = log(4)
log(c+1) = log(2) (and not because 4/2=2)
c = 1
A. To find the area of region R, we need to calculate the integral of the function y = 1/(sqrt(x+1)) from x = 0 to x = 3.
∫[0,3] 1/(sqrt(x+1)) dx
To solve this integral, we can use u-substitution with u = x + 1, which gives us du = dx.
When x = 0, u = 1 and when x = 3, u = 4.
∫[1,4] 1/√u du
Using the power rule for integration, we have:
∫[1,4] u^(-1/2) du
To integrate u^(-1/2), we can add 1 to the exponent and divide by the new exponent:
∫[1,4] u^(1/2) / (1/2) du
Simplifying, we have:
2 ∫[1,4] u^(1/2) du
Integrating u^(1/2), we get:
2 * (2/3)u^(3/2) evaluated from 1 to 4
2 * (2/3)(4^(3/2) - 1^(3/2))
2 * (2/3)(8 - 1)
2 * (2/3)(7)
28/3
Therefore, the area of region R is 28/3 square units.
B. To find the volume of the solid formed when region R is revolved about the x-axis, we'll use the disk method.
The radius of each disk is given by y, which is 1/(sqrt(x+1)).
To calculate the volume, we need to evaluate the integral:
V = π∫[0,3] [1/(sqrt(x+1))]^2 dx
Simplifying,
V = π∫[0,3] [1/(x+1)] dx
Using the substitution u = x+1, du = dx, the integral becomes:
V = π∫[1,4] [1/u] du
V = π ln(u) | [1,4]
V = π [ ln(4)-ln(1) ]
V = π ln(4)
Therefore, the volume of the solid formed when region R is revolved about the x-axis is π ln(4) cubic units.
C. To find the x-value where the plane intersects the x-axis and divides the solid into two equal volumes, we need to find the x-value that corresponds to half the volume.
Since we know the volume is given by V = π ln(4), we can set up the equation:
π ln(4)/2 = π ln(4x)/2
Cancelling the common factors, we have:
ln(4)/2 = ln(4x)/2
Taking the exponential of both sides:
e^(ln(4)/2) = e^(ln(4x)/2)
Simplifying, we get:
√(4) = √(4x)
2 = 2√x
√x = 1
x = 1
Therefore, the x-value where the plane intersects the x-axis and divides the solid into two equal volumes is x = 1.
A. To find the area of region R, we can integrate the function y = 1/(sqrt(x+1)) over the given bounds.
The region R is bounded by the x-axis, y-axis, x = 3, and y = 1/(sqrt(x+1)). This means we need to find the values of x where y = 0 and y = 1/(sqrt(x+1)) intersect.
Setting y = 0, we have:
0 = 1/(sqrt(x+1))
0 = 1
There are no solutions to this equation, which means that the function y = 1/(sqrt(x+1)) does not intersect the x-axis in the given range.
To find the area of region R, we need to integrate the function y = 1/(sqrt(x+1)) from x = 0 to x = 3:
Area = ∫[0, 3] 1/(sqrt(x+1)) dx
To evaluate this integral, we can make a substitution. Let u = x + 1, then du = dx. When x = 0, u = 1, and when x = 3, u = 4.
So the integral becomes:
Area = ∫[1, 4] 1/(sqrt(u)) du
This is equal to 2√u evaluated from 1 to 4:
Area = 2(√4 - √1)
Area = 2(2 - 1)
Area = 2 square units.
Therefore, the area of region R is 2 square units.
B. To find the volume of the solid formed when the region R is revolved about the x-axis, we can use the disk method.
The volume of the solid is given by the integral:
Volume = ∫[0, 3] π(1/(sqrt(x+1)))^2 dx
Simplifying this expression, we have:
Volume = ∫[0, 3] π/(x+1) dx
To evaluate this integral, we can use the natural logarithm function. Applying the fundamental theorem of calculus, we get:
Volume = π ln(x+1) evaluated from 0 to 3
Volume = π(ln(3+1) - ln(0+1))
Since ln(1) = 0, this becomes:
Volume = πln(4).
Therefore, the volume of the solid formed when region R is revolved about the x-axis is πln(4) cubic units.
C. The x-value where the plane intersects the x-axis can be found by dividing the given volume by 2 and then finding the corresponding x-value.
Volume of each solid formed = (1/2) * Volume from part B
Let V1 be the volume of the first solid, and V2 be the volume of the second solid.
(1/2) * πln(4) = V1 = V2
To find the x-value, we can reverse the process of finding the volume. We need to solve the equation:
(1/2) * πln(x+1) = (1/2) * πln(4)
Simplifying, we have:
ln(x+1) = ln(4)
Since ln is a one-to-one function, this implies:
x + 1 = 4
Solving for x, we find:
x = 4 - 1
x = 3.
Therefore, the x-value where the plane intersects the x-axis is 3.