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Volume of a Segment of a Paraboloid The region bounded by \(y=r^{2}-x^{2}, y=0,\) and \(x=0\) is revolved about the \(y\) -axis to form a paraboloid. A hole, centered along the axis of revolution, is drilled through this solid. The hole has a radius \(k, 0

Short Answer

Expert verified
The volume of the resulting ring can be found by subtracting the volume of the drilled hole from the volume of the original shape, and can be done by integrating with respect to both \(x\) and \(y\). Fully solving the problem would require calculation of these integrals.

Step by step solution

01

Setting up the Integral for Part a

When integrating with respect to \(x\), the volume \(V\) of the resulting ring can be defined as the difference between the volume of the original shape and the volume of the drilled hole. The volume of the original shape is given by the formula for the volume of revolution, \(V = \pi \int_0^r (r^2 - x^2)^2 dx\). The volume of the drilled hole is given by \(V_h = \pi \int_0^k (k^2) dx\). The volume of the resulting ring is then \(V - V_h\).
02

Calculating the Integral for Part a

To calculate the integral for part a, first calculate the integral for the volume of the original shape and the volume of the drilled hole separately. After calculating these integrals, subtract the volume of the drilled hole from the volume of the original shape to get the volume of the ring.
03

Setting up the Integral for Part b

When integrating with respect to \(y\), use the same concept as in step 1. The volume of the original shape is given by the formula for the volume of revolution, \(V = \pi \int_0^r y dx\), the volume of the drilled hole is \(V_h = \pi \int_0^k y dx\). Again, the volume of the resulting ring is \(V - V_h\).
04

Calculating the Integral for Part b

To calculate the integral for part b, follow the same steps as in step 2. Calculate the integral for the volume of the original shape and the volume of the hole separately, then subtract the volume of the hole from that of the original shape to get the volume of the ring.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integral Calculus
Integral calculus is a fundamental part of mathematics that deals with the accumulation of quantities and the area under curves. When applied to find volumes, integral calculus is a powerful tool that can calculate the space occupied by almost any three-dimensional object. To find the volume of an irregular shape like a segment of a paraboloid, we set up a definite integral that represents the summation of an infinite number of infinitely thin slices of the shape.

The concept of integrating with respect to a given variable, such as \(x\) or \(y\), implies summing these slices along the respective axis. In the exercise, the difference in volumes—between the original paraboloid and the drilled hole—is calculated by setting up and evaluating definite integrals. This requires the boundaries of integration, which are provided by the limits of the region being revolved, and understanding the relationship between the variables, such as the equation \(y = r^2 - x^2\), which describes the parabola in this case.

The volume calculations become a subtractive process, where we evaluate the integrals separately and then find the volume difference to get the desired volume of the ring created by drilling a hole through the paraboloid.
Volume of Revolution
The volume of revolution is an application of integral calculus used to compute the volume of a three-dimensional solid formed by rotating a two-dimensional region around an axis. In our case, the axis of rotation is the \(y\)-axis, and the two-dimensional region is bounded by the parabolic equation. For calculating the volume of such solids, we apply an approach called the 'disk method', where we imagine slicing the solid into a series of thin disks or washers.

The volume of each disk is approximated as \(\text{Volume of the disk} = \text{Area of circular face} \times \text{thickness}\). The area of the circular face is \(\text{Area} = \text{Radius}^2 \times \text{Pi}\), where the radius is expressed in terms of our variables \(x\) and \(y\). By adding up the volumes of all these disks along the axis of rotation using integration, we find the total volume of the solid. The integration boundaries for the exercise are dictated by the region we are rotating, in this case from \(0\) to \(r\) along the \(x\)-axis.
Method of Disks
The method of disks is a technique within the volume of revolution that simplifies the process of finding volumes of solids with circular cross-sections perpendicular to the axis of rotation. Imagine taking a solid and slicing it into a stack of disks, much like slices of bread. Each disk's thickness represents an infinitely small segment of the solid, and the volume of each disk can be calculated by its cross-sectional area—essentially the face of the slice.

When using this method, the area of the disk is dependent on the function that defines the boundary of our solid. For the paraboloid mentioned in the exercise, the boundary is described by \(y = r^2 - x^2\). Hence, the radius of each disk is a function in terms of \(x\), and the area is \(\text{Area} = \text{Pi} \times (r^2 - x^2)^2\). Once we have the area as a function of \(x\), we can integrate along the axis of revolution, which gives the total volume. This method is very visually intuitive, as one can imagine adding an infinitesimally thin disk onto another to build up the entire volume of the solid.

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