Question

Let \(R\) be the region bounded by the upper half of the ellipse \(x^{2} / 2+y^{2}=1\) and the parabola \(y=x^{2} / \sqrt{2}\) a. Find the area of \(R\). b. Which is greater, the volume of the solid generated when \(R\) is revolved about the \(x\) -axis or the volume of the solid generated when \(R\) is revolved about the \(y\) -axis?

Short answer

Question: Find the area of region R bounded by the upper half of the ellipse \(x^2/2+y^2=1\) and the parabola \(y=x^2/\sqrt{2}\). Then, find the volume of the solid generated when the region R is revolved about the x-axis and the y-axis, respectively, and determine which volume is greater. Answer: Calculate the area R using integration, and use the disk method to find the volumes of revolution about the x-axis and y-axis. Compare the volumes to determine which one is greater.

Step-by-step solution

Rewrite the ellipse and parabola equations

We first need to isolate y in both equation for the ellipse and parabola so that we can calculate the area R. Ellipse equation: \(\frac{x^2}{2}+y^2=1\). Rewrite the equation to get \(y^2 = 1-\frac{x^2}{2}\), thus, \(y = \sqrt{1-\frac{x^2}{2}}\). Parabola equation: \(y=x^2/\sqrt{2}\).

Find the points of intersection

We need to find the intersection points between the ellipse and parabola to determine the limits of integration. Set the two equations equal to each other: \(\sqrt{1-\frac{x^2}{2}}=x^2/\sqrt{2}\). Square both sides: \(1-\frac{x^2}{2} =\frac{x^4}{2}\). To find the roots, set the equation to zero: \(x^4-x^2+2=0\) which is a quadratic equation in disguise. Define \(u=x^2\), then the equation becomes \(u^2-u+2=0\). Since there are no real roots, we need to find the intersection points graphically which will give us \(x=-\sqrt{2}/2\) and \(x=\sqrt{2}/2\).

Calculate the area of R

Use integration to find the area of region R: \(R=\int_{-\sqrt{2}/2}^{\sqrt{2}/2}(\sqrt{1-\frac{x^2}{2}}-\frac{x^2}{\sqrt{2}})dx\).

Find the volume of revolution about the x-axis and the y-axis

We will use the disk method to find the volume of revolution. For the x-axis: we integrate from \(-\sqrt{2}/2\) to \(\sqrt{2}/2\): \(V_x=\pi\int_{-\sqrt{2}/2}^{\sqrt{2}/2}(y^2_dx)=\pi\int_{-\sqrt{2}/2}^{\sqrt{2}/2}\left((1-\frac{x^2}{2})-\frac{x^4}{2}\right)dx\). For the y-axis: we integrate from 0 to 1 : \(V_y=2\pi\int_{0}^{1}(x^2+y^2/2)dy=2\pi\int_{0}^{1}\left(\frac{1}{2}y+\frac{y^2}{2}\right)dy\).

Compare the volumes

Once you have calculated \(V_x\) and \(V_y\), compare the two volumes to determine which one is greater.

Key concepts

Ellipse

An ellipse is a shape that looks like a flattened circle. It has two axes: a longer one called the major axis, and a shorter one called the minor axis. In the exercise, we are working with the upper half of an ellipse described by the equation \(\frac{x^2}{2} + y^2 = 1\). To simplify calculations, you can solve for \(y\), getting \(y = \sqrt{1 - \frac{x^2}{2}}\).

• The center of this ellipse is at the origin (0,0).
• The semi-major axis is oriented along the x-axis.
• The semi-minor axis is oriented along the y-axis.

Understanding the shape and orientation of an ellipse is essential for solving problems related to its boundaries, like finding areas and volumes when it is revolved around axes.

Parabola

A parabola is a U-shaped curve that can open upwards, downwards, left, or right. In this scenario, we're dealing with a parabola described by the function \(y = \frac{x^2}{\sqrt{2}}\). Here, the parabola opens upwards because, as \(x\) increases or decreases, \(y\) becomes larger too:

• The vertex of this parabola is at the origin (0,0).
• It is symmetric about the y-axis.
• Parabolas are often used in problems involving quadratic functions.

Recognizing how a parabola behaves and intersects with other shapes, such as an ellipse, is key to solving the problem of finding intersections and calculating the area between these curves.

Area of Region

Finding the area of the region bounded by curves involves determining the intersection points and setting up a proper integral. Here, we're interested in the area \(R\) between the upper half of an ellipse and a parabola. First, we establish the limits of integration by solving the equation \(\sqrt{1-\frac{x^2}{2}} = \frac{x^2}{\sqrt{2}}\), representing the height differences of these curves as functions of \(x\). The solutions, \(x = -\frac{\sqrt{2}}{2}\) and \(x = \frac{\sqrt{2}}{2}\), define the interval for integration:

• Set up the integral with these limits: \(R = \int_{-\sqrt{2}/2}^{\sqrt{2}/2}\left(\sqrt{1-\frac{x^2}{2}} - \frac{x^2}{\sqrt{2}}\right)dx\).
• Calculate the area by evaluating this integral.

Grasping the methodology of setting these boundaries and calculating the integral helps to understand how various curves form enclosed regions.

Integration

Integration is a fundamental concept in calculus, used to calculate areas, volumes, and other quantities. In this problem, integration is used to find the area of the region \(R\) and the volumes generated by revolving \(R\) around the x-axis and y-axis.For the area, we use the definite integral over an interval determined by the intersection points:

• Integrate the function \(\sqrt{1-\frac{x^2}{2}} - \frac{x^2}{\sqrt{2}}\) over the interval \([-\sqrt{2}/2, \sqrt{2}/2]\).

For volumes, integration is used in the disk method:

• Volume around x-axis: \(V_x = \pi \int_{-\sqrt{2}/2}^{\sqrt{2}/2} \left((1-\frac{x^2}{2}) - \frac{x^4}{2}\right)dx\)
• Volume around y-axis: \(V_y = 2\pi \int_{0}^{1} \left(\frac{1}{2}y + \frac{y^2}{2}\right)dy\)

Mastering these integration techniques allows a deeper comprehension of how calculus can calculate areas and volumes related to intersections of different curves.