Question

Which bowl holds more water if it is filled to a depth of 4 units? \(\cdot\) The paraboloid \(z=x^{2}+y^{2},\) for \(0 \leq z \leq 4\) \(\cdot\) The cone \(z=\sqrt{x^{2}+y^{2}},\) for \(0 \leq z \leq 4\) \(\cdot\) The hyperboloid \(z=\sqrt{1+x^{2}+y^{2}},\) for \(1 \leq z \leq 5\)

Short answer

Answer: The hyperboloid can hold the most water when filled to a depth of 4 units.

Step-by-step solution

Paraboloid Volume Calculation

First, we need to calculate the volume of the paraboloid. This can be done using the polar coordinate system and then integrating using the limits to find the volume. We can also make use of symmetry. In polar coordinates, \(x^2 + y^2 = r^2\). Hence, the equation of the paraboloid is given by \(z = r^2\). Now we can find the volume of the paraboloid using the following integral: $$ V_1 = \int_{0}^{2\pi} \int_{0}^{\sqrt{4}} \int_{0}^{4} r^3 dr d\theta dz $$

Cone Volume Calculation

To calculate the volume of the cone, we use its equation in polar coordinates: \(z = r\). Now, we can find the volume of the cone using the following integral: $$ V_2 = \int_{0}^{2\pi} \int_{0}^{4} \int_{0}^{4} r dr d\theta dz $$

Hyperboloid Volume Calculation

For the hyperboloid, we first need to find the value of \(r\) at which the function intersects the plane \(z=5\). The function is given by \(z = \sqrt{1 + x^2 + y^2}\), which in polar coordinates becomes \(z = \sqrt{1 + r^2}\). Setting \(z=5\), we get \(r^2 = 24\). To find the volume, we integrate using the following integral: $$ V_3 = \int_{0}^{2\pi} \int_{0}^{\sqrt{24}} \int_{1}^{5} r (\sqrt{1+r^2}-1) dr d\theta dz $$

Finding and Comparing Volumes

After evaluating the above integrals, we compare the volumes of the three shapes. The one with the largest volume will hold the most water. By calculating the integrals above, we get \( V_1=\frac{32\pi}{3} \), \( V_2=16\pi \) and \( V_3=\frac{278\pi}{3}\). Comparing the volumes of these three bowls, we find that the hyperboloid has the largest volume, and therefore can hold the most water when filled to a depth of 4 units.

Key concepts

Volume Calculation

When faced with a problem involving finding which bowl shape holds more water, it’s crucial to understand how to calculate volumes of different shapes.
For this, we rely on mathematical formulas for their volumes, often using calculus.
In our example, we're dealing with a paraboloid, a cone, and a hyperboloid, each of which can be expressed with specific equations.

To determine which of these holds more water, we look at their respective volume using integration.
Each shape has unique boundaries which are defined by various limits and equations.
These boundaries help determine how much space the shape actually occupies in 3D space if filled up to a certain depth.

We use integration because it allows us to add up contributions of infinite small elements to find the actual volume.

• The paraboloid follows the equation: \(z = r^2\) within 0 to 4 units in the \(z\)-direction.
• The cone is represented by \(z = r\), with a maximum height of 4 units.
• The hyperboloid requires special care. The surface defined by \(z = \sqrt{1 + x^2 + y^2}\) starts above the base at 1 unit and ends at 5 units, defining the height.

By inserting these parameters into the integration process, and computing, each shape's volume is estimated and compared to determine which holds more.

Polar Coordinates

Polar coordinates offer a simpler way of representing two-dimensional figures like circles and curves, often simplifying the processing of 3D solids.

In this exercise, we use polar coordinates to express shapes like paraboloids and cones which are symmetrical around their axis.

Polar coordinates are represented by the variables \(r\) and \(\theta\).

• \(r\) is the radius or distance from the origin to a point.
• \(\theta\) is the angle from the positive \(x\)-axis to the line that connects the origin and the point.

In cartesian coordinates, a point is described by \( (x, y) \) but in polar, it is \((r, \theta)\).
This conversion gets really useful in equations of rotations and radial symmetry.

The advantage of polar coordinates lies in their ability to simplify the setup of integrals when dealing with circular or rotational symmetry.
Working within the polar coordinate system, we replace \(x\) and \(y\) with \(r\cos(\theta)\) and \(r\sin(\theta)\). This makes integration in these contexts easier, especially when rotating around the origin, as we have with our shapes.

Integration Techniques

Integrals allow us to sum up small slices or sections of an object, which together can evaluate to a full volume.

For 3D objects like the ones in our problem, triple integrals can effectively calculate volumes by summing all little volumes or 'slices'.
Let's focus on integrating with limits since they are the boundaries of the volume.
To solve such problems, we utilize the follows steps:

• Convert necessary equations to polar form.
• Identify each shape's boundaries and determine limits for integration.
• Set up integrals that represent the sum of all small radius slices across 3D space until the maximum boundary.

These steps outline how the integration of paraboloid, cone, and hyperboloid are achieved.
By following these integrals:
- For the paraboloid: Limits are set in polar to include radius and angle, allowing for symmetry utilization. We calculate from the outer edge inward. - For the cone: Simpler equations make for easier setup, allowing easy limits within zero to maximum height determined by the cone's steep slope. - The more complex hyperboloid: It requires moving away from a simple base and understanding that its form starts not from zero. Mastery over these techniques is essential as it sets apart basic and advanced calculus applications.