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. To find the area of region R, we need to compute the integral of the function that represents the curve y = 1/(sqrt(x+1)) over the given range. We can express the area as:
A = ∫[0 to 3] 1/(sqrt(x+1)) dx
First, let's simplify the integral by making a substitution. Let u = x + 1.
This implies du = dx, and when x = 0, u = 1, and when x = 3, u = 4.
Therefore, our integral becomes:
A = ∫[1 to 4] 1/(sqrt(u)) du
Using the power rule for integrals, we can simplify this to:
A = 2√u | [1 to 4]
Evaluating this expression at the limits, we have:
A = 2√4 - 2√1
A = 2(2) - 2(1)
A = 4 - 2
A = 2 square units
So, the area of region R is 2 square units.
B. To find the volume of the solid formed when region R is revolved about the x-axis, we can use the disk method.
The volume V of the solid can be calculated by integrating the area of each infinitesimally small disk formed by rotating the function 1/(sqrt(x+1)) around the x-axis from x = 0 to x = 3.
V = ∫[0 to 3] π[1/(sqrt(x+1))]^2 dx
Simplifying the expression:
V = π∫[0 to 3] 1/(x+1) dx
We can now evaluate this integral by taking the natural logarithm (ln) of the absolute value of x + 1:
V = πln|x+1| | [0 to 3]
V = πln(4) - πln(1)
V = πln(4)
So, the volume of the solid formed when region R is revolved about the x-axis is πln(4) cubic units.
C. To find the x-value where the plane intersects the x-axis and divides the solid formed in part B into two equal volumes, we need to set up and solve an equation.
Let's assume the plane intersects the x-axis at x = a. Then, we have two volumes, V1 and V2, such that:
V1 = V2
Using the disk method, we can express these volumes as:
V1 = π∫[0 to a] 1/(x+1) dx
V2 = π∫[a to 3] 1/(x+1) dx
Setting V1 equal to V2:
π∫[0 to a] 1/(x+1) dx = π∫[a to 3] 1/(x+1) dx
Now, we can solve this equation:
∫[0 to a] 1/(x+1) dx = ∫[a to 3] 1/(x+1) dx
By taking the antiderivative of both sides, we get:
ln(a+1) - ln(0+1) = ln(3+1) - ln(a+1)
Since ln(0+1) = ln(1) = 0, we have:
ln(a+1) = ln(4) - ln(a+1)
To eliminate the logarithms, we can exponentiate both sides:
e^(ln(a+1)) = e^(ln(4) - ln(a+1))
This simplifies to:
a + 1 = 4/(a+1)
Now, we can solve this quadratic equation:
(a + 1)^2 = 4
Expanding and simplifying:
a^2 + 2a + 1 = 4
a^2 + 2a - 3 = 0
Factoring:
(a + 3)(a - 1) = 0
This gives us two potential solutions: a = -3 and a = 1.
However, since a represents the x-value, it cannot be negative in this context. Therefore, the x-value where the plane intersects the x-axis and divides the solid into two equal volumes is a = 1.
To find the area of region R, we need to calculate the definite integral of the curve.
A. Finding the area of region R:
We can see that the curve y = 1 / √(x + 1) intersects the x-axis at x = 0 and x = 3.
The area of region R can be calculated as follows:
Area = ∫(0 to 3) y dx
Let's evaluate the integral:
Area = ∫(0 to 3) 1/√(x + 1) dx
To evaluate this integral, we can make a substitution. Let's substitute u = x + 1, so du = dx.
Area = ∫(0 to 3) 1/√u du
Now the integral becomes:
Area = ∫(1 to 4) 1/√u du
Using the power rule of integration, this can be represented as:
Area = 2√u |(1 to 4)
Evaluating this:
Area = 2(√4 - √1)
= 2(2 - 1)
= 2 square units
Therefore, the area of region R is 2 square units.
B. Finding the volume of the solid formed when region R is revolved about the x-axis:
To find the volume, we need to use the disk method. The volume of the solid can be calculated using the formula:
V = π ∫(a to b) [f(x)]^2 dx
In this case, the function is f(x) = 1/√(x + 1), and the interval is (0 to 3).
Therefore, the integral becomes:
V = π ∫(0 to 3) [1/√(x + 1)]^2 dx
V = π ∫(0 to 3) 1/(x + 1) dx
Using the u-substitution method, let's substitute u = x + 1, so du = dx.
V = π ∫(1 to 4) 1/u du
V = π ln|u| | (1 to 4)
V = π(ln|4| - ln|1|)
V = π(ln4)
Therefore, the volume of the solid formed when region R is revolved about the x-axis is π(ln4).
C. Finding the x-value where the plane intersects the x-axis:
To divide the solid into two equal volumes, we need to find the value of x where the two volumes are equal.
Let's assume the plane intersects the x-axis at the point (x, 0).
The volume of the solid on the left side of this plane is given by:
V1 = π ∫(0 to x) [1/√(t + 1)]^2 dt
Similarly, the volume of the solid on the right side of this plane is given by:
V2 = π ∫(x to 3) [1/√(t + 1)]^2 dt
To find the point of intersection, we need to equate V1 and V2:
π ∫(0 to x) [1/√(t + 1)]^2 dt = π ∫(x to 3) [1/√(t + 1)]^2 dt
Cancelling out π and simplifying the fractions:
∫(0 to x) [1/√(t + 1)]^2 dt = ∫(x to 3) [1/√(t + 1)]^2 dt
∫(0 to x) 1/(t + 1) dt = ∫(x to 3) 1/(t + 1) dt
Now, we can evaluate these two integrals:
ln|t + 1| |(0 to x) = ln|t + 1| |(x to 3)
ln|x + 1| - ln|1| = ln|3 + 1| - ln|x + 1|
ln|x + 1| = ln4 - ln|x + 1|
Taking the exponential of both sides:
|x + 1| = (4)/(|x + 1|)
|x + 1|^2 = 4
Now, solving for x:
(x + 1)^2 = 4
x + 1 = ±2
x = -1 ± 2
Therefore, the two x-values where the plane intersects the x-axis are x = -3 and x = 1.
I hope this helps! Let me know if you have any further questions.