Problem 6 - University Calculus: Early Transcendentals

Question

Let \(D\) be the region bounded by the paraboloid \(z=x^{2}+y^{2}\) and the plane \(z=2 y\) Write triple iterated integrals in the order \(d z d x d y\) and \(d z d y d x\) that give the volume of \(D .\) Do not evaluate either integral.

Step-by-Step Solution

Answer

Volume of region given by integrals: \[\int_0^2 \int_{-\sqrt{1-(y-1)^2}}^{\sqrt{1-(y-1)^2}} \int_{x^2+y^2}^{2y} 1\, dz\, dx\, dy\] and \[\int_{-1}^1 \int_{1-\sqrt{1-x^2}}^{1+\sqrt{1-x^2}} \int_{x^2+y^2}^{2y} 1\, dz\, dy\, dx.\]

Step 1: Understanding the Intersection

To find the intersection of the paraboloid and the plane, set their equations equal: \(x^2 + y^2 = 2y\). This simplifies to \(x^2 + (y-1)^2 = 1\), which is a circle with radius 1 and center at \((0, 1)\).

Step 2: Determining the Bounds for \(z\)

The volume region \(D\) is bounded above by \(z = 2y\) (the plane) and below by \(z = x^2 + y^2\) (the paraboloid). Thus, for a fixed \(x\) and \(y\), \(z\) ranges from \(z = x^2 + y^2\) to \(z = 2y\).

Step 3: Setting the Bounds for \(x\) and \(y\) in \(dz\,dx\,dy\) Order

For \(dz\,dx\,dy\), consider the circle in the \(xy\)-plane which is \((0, 1)\) centered. For a given \(y\), \(x\) ranges from \(-\sqrt{1 - (y - 1)^2}\) to \(\sqrt{1 - (y - 1)^2}\). \(y\) ranges from 0 to 2 to cover the circle.

Step 4: Setting up the Iterated Integral with Order \(dz\,dx\,dy\)

The iterated integral is: \[\int_0^2 \int_{-\sqrt{1 - (y - 1)^2}}^{\sqrt{1 - (y - 1)^2}} \int_{x^2 + y^2}^{2y} 1\, dz\, dx\, dy.\]

Step 5: Setting the Bounds for \(x\) and \(y\) in \(dz\,dy\,dx\) Order

For \(dz\,dy\,dx\), switch the integration order. For given \(x\), \(y\) ranges from solving \(y - 1 = \pm \sqrt{1 - x^2}\), so \(1 - \sqrt{1 - x^2} \leq y \leq 1 + \sqrt{1 - x^2}\). \(x\) ranges from -1 to 1.

Step 6: Setting up the Iterated Integral with Order \(dz\,dy\,dx\)

The iterated integral is: \[\int_{-1}^1 \int_{1 - \sqrt{1 - x^2}}^{1 + \sqrt{1 - x^2}} \int_{x^2 + y^2}^{2y} 1\, dz\, dy\, dx.\]

Step 7: Final Note

Both integrals ultimately provide the volume of the region \(D\), although their orders and specific variable bounds differ.

Key Concepts

Paraboloid

A paraboloid is a three-dimensional geometric shape that resembles a parabolic curve extended along an axis. In this exercise, the paraboloid in question is defined by the equation \(z = x^2 + y^2\). To visualize, imagine a shape like an open bowl or an umbrella, with the tip pointing downwards along the \(z\)-axis.

• It opens upwards along the \(z\)-axis, capturing any point above the origin based on the squared terms.
• Any cross-section parallel to the \(xy\)-plane is a circle, making it a rotational paraboloid.

The intersection with a plane, such as \(z = 2y\), involves aligning these equations to find a common region. For this exercise, the intersection creates a boundary that aids in volume calculation.

Iterated Integrals

Iterated integrals allow us to compute the volume under a surface encompassed by multiple variables. They are executed one variable at a time, in a specific order, simplifying the calculations. In this exercise, we write two different iterated integrals for volume:

• The first is in \(dz\, dx\, dy\) order. Integration starts with \(z\), followed by \(x\), and finally \(y\).
• The second uses the \(dz\, dy\, dx\) order. It begins with \(z\), then \(y\), and ends with \(x\).

By following these orders, we can establish bounds for each integral effectively. The integral with respect to \(z\) generally bounds the height of the solid, while the subsequent integrals determine the other dimensions along the \(x\) and \(y\) axes respectively. This meticulous approach ensures the entire volume of the desired structure is captured.

Volume Calculation

Calculating the volume of a region bounded by complex surfaces involves using triple integrals. These integrals sum infinitely small volumes throughout the space to find the total volume. For this specific problem:

• The region \(D\) is bounded below by the paraboloid \(z = x^2 + y^2\) and above by the plane \(z = 2y\).
• The determined bounds for \(z\) are \(z = x^2 + y^2\) to \(z = 2y\), based on the intersection of the paraboloid and plane.

Using the derived iterated integrals:

• In \(dz\, dx\, dy\) order, \(x\) ranges from \(-\sqrt{1-(y-1)^2}\) to \(\sqrt{1-(y-1)^2}\) and \(y\) ranges from 0 to 2.
• In \(dz\, dy\, dx\) order, \(y\) ranges from \(1 - \sqrt{1-x^2}\) to \(1 + \sqrt{1-x^2}\), and \(x\) spans from -1 to 1.

These integrals are set up but not evaluated, indicating the structure required for computation, essential when multiple calculation paths are available in math problems. By choosing convenient integration orders, we can often simplify the computational complexity.