Question

Volumes of solids Find the volume of the following solids. The region bounded by $$y=\frac{x}{x+1},\( the \)x\( -axis, and \)x=4$$ is revolved about the \(x\) -axis.

Short answer

Question: Calculate the volume of the solid generated by revolving the region bounded by the function \(y=\frac{x}{x+1}\), the \(x\)-axis, and \(x=4\) around the \(x\)-axis using the disk method. Answer: The volume of the solid is given by \(V = \pi (4 - 2\ln{5} + 1/5)\).

Step-by-step solution

Write down the function and integration interval

We are given the function \(y=\frac{x}{x+1}\). The region we are revolving around the \(x\)-axis is bounded by the \(x\)-axis and \(x=4\). Therefore, our integration interval is from 0 to 4.

Determine the volume formula using the disk method

For the disk method, the volume is found by integrating the area of each disk in the given interval. The formula to find the volume is: $$V = \pi \int_a^b [r(x)]^2 dx$$ Where \(V\) is the volume, \(r(x)\) is the radius, and \(a\) and \(b\) are the limits of integration.

Calculate the radius of each disk

The radius of each disk is equal to the value of the function \(y\) at a given \(x\). In this case, \(r(x) = y = \frac{x}{x+1}\).

Plug radius into volume formula and set up the integral

Using the radius we found in Step 3, we can now set up the integral for the volume formula: $$V = \pi \int_0^4 \left(\frac{x}{x+1}\right)^2 dx$$

Simplify the integral

Simplify the integral before evaluating it: $$V = \pi \int_0^4 \frac{x^2}{(x+1)^2} dx$$

Evaluate the integral

To evaluate the integral, we can use the substitution method. Let $$u = x+1$$ Then $$du = dx$$ and $$x = u-1$$ Our new integral becomes: $$V = \pi \int_1^5 \frac{(u-1)^2}{u^2} du$$ Expanding the numerator and simplifying, we get: $$V = \pi \int_1^5 \frac{u^2 - 2u + 1}{u^2} du$$ Split the integral into separate fractions: $$V = \pi \int_1^5 (1 - \frac{2}{u} + \frac{1}{u^2}) du$$ Now, integrate each term: $$V = \pi \left[u - 2\ln{(u)} - \frac{1}{u}\right]_1^5$$

Compute the volume

Evaluate the expression by plugging in the limits of integration: $$V = \pi [(5 - 2\ln{5} - 1/5) - (1 - 2\ln{1} - 1)]$$ Simplify further: $$V = \pi (4 - 2\ln{5} + 1/5)$$ So, the volume of the solid generated by revolving the region bounded by the function \(y=\frac{x}{x+1}\), the \(x\)-axis, and \(x=4\) around the \(x\)-axis is: $$V = \pi (4 - 2\ln{5} + 1/5)$$

Key concepts

Disk Method

The disk method is a technique used to find the volume of a solid of revolution. It works by slicing the solid into thin disks perpendicular to the axis of revolution.
Each disk resembles a cylinder with a very small height. This method is particularly useful when the solid is revolved around the x-axis or y-axis.

• To employ the disk method, identify a function that emits the boundary of the region being revolved.
• Each of these disks has a radius equal to the distance from the axis of revolution to the function line.
• The volume of the solid is then calculated by integrating the area of these disks over the given interval.

In this problem, the function provided is the boundary of the region that will be revolved around the x-axis. With it, we can find each disk's radius and set up our integral to find the volume.

Integration

Integration is a fundamental concept in calculus, used to find areas, volumes, central points, and many useful things. When using the disk method, integration helps us add up an infinite number of infinitesimally small disks.
As a result, it provides us with the total volume of the solid of revolution.

• The integral in the disk method calculates the infinite sum: it accumulates the area of all disks into a single numerical value.
• This process involves defining a suitable definite integral and solving it, either by applying integration techniques or substitution if simplification is needed.
• In our example, the function was squared and integrated over the interval from 0 to 4.

Solving the integral requires breaking it down into manageable parts or simpler functions, facilitating the calculations involved in finding the volume.

Substitution Method

The substitution method is a useful technique in integration when the integrand is complex. It involves changing variables to simplify the integration process.
By substituting a part of the original integrand with a single variable, the new integral often becomes more straightforward to solve.

• This technique requires a change of variable that transforms the integrand into a simpler form.
• Once the integral is expressed in terms of the new variable, we perform the integration.
• Substitute back to the original variable once integration is complete to find the result in terms of the original limits.

In the problem at hand, substituting with 'u = x+1' made the computation smoother by producing an easier integrand to manage.

Revolution Around the X-axis

Revolution around the x-axis occurs when a region in the xy-plane is revolved around the x-axis to form a three-dimensional solid.
This concept is foundational in finding volumes of such solids with the disk or washer methods.

• When a shape in the plane is revolved around the x-axis, each point on the shape traces out a three-dimensional form, often a cylinder or paraboloid.
• The volume of the resulting solid is calculated using integration methods, such as the disk or washer method.
• In our scenario, integrating around the x-axis allowed us to compute the volume of a solid formed by revolving the defined region.

This approach helps in visualizing and computing volumes of solids with well-defined boundaries and integrals, essential for solutions in calculus.