applications of integrals: x=y^2, x=1-y^2

The given curves are:

1. \( x = y^2 \)
2. \( x = 1 - y^2 \)

### Step 1: Understand the curves

- The curve \( x = y^2 \) is a parabola that opens to the right, with its vertex at the origin.
- The curve \( x = 1 - y^2 \) is also a parabola, but it opens to the left, and its vertex is at \( (1, 0) \).

### Step 2: Find the points of intersection

To find the points where these curves intersect, set the equations equal to each other:
\[
y^2 = 1 - y^2
\]
\[
2y^2 = 1
\]
\[
y^2 = \frac{1}{2}
\]
\[
y = \pm \frac{1}{\sqrt{2}}
\]

Thus, the curves intersect at \( y = \pm \frac{1}{\sqrt{2}} \).

### Step 3: Set up the integral

Now, we'll find the area between the curves. The limits of integration will be from \( y = -\frac{1}{\sqrt{2}} \) to \( y = \frac{1}{\sqrt{2}} \). The area is given by the integral of the difference between the two functions for \( x \).

For any \( y \), the right curve is \( x = 1 - y^2 \) and the left curve is \( x = y^2 \), so the area \( A \) is:

\[
A = \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} [(1 - y^2) - y^2] \, dy
\]
\[
A = \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} (1 - 2y^2) \, dy
\]

### Step 4: Solve the integral

We can now compute the integral:

\[
A = \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} 1 \, dy - 2 \int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} y^2 \, dy
\]

1. The first integral is straightforward:
\[
\int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} 1 \, dy = 2 \times \frac{1}{\sqrt{2}} = \sqrt{2}
\]

2. The second integral involves \( y^2 \):
\[
\int_{-\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}} y^2 \, dy = 2 \times \int_0^{\frac{1}{\sqrt{2}}} y^2 \, dy
\]
\[
= 2 \times \left[ \frac{y^3}{3} \right]_0^{\frac{1}{\sqrt{2}}}
\]
\[
= 2 \times \frac{1}{3} \times \left( \frac{1}{\sqrt{2}} \right)^3 = 2 \times \frac{1}{3} \times \frac{1}{2\sqrt{2}} = \frac{1}{3\sqrt{2}}
\]

### Step 5: Combine results

Now, subtract the second result from the first:

\[
A = \sqrt{2} - 2 \times \frac{1}{3\sqrt{2}} = \sqrt{2} - \frac{2}{3\sqrt{2}}
\]

This simplifies to:

\[
A = \frac{3\sqrt{2}}{3} - \frac{2}{3\sqrt{2}} = \frac{3 - 2}{3\sqrt{2}} = \frac{1}{3\sqrt{2}}
\]

Thus, the area between the curves is \( \frac{1}{3\sqrt{2}} \).

---

### Applications of Integrals:

1. **Area between curves:** As calculated above, integrals are useful for finding areas between curves.
2. **Volume of solids:** By rotating curves around an axis, integrals can calculate the volume of the resulting solid.
3. **Work done by a force:** If a force varies over a distance, integrals compute the total work done.
4. **Center of mass:** The integral helps find the center of mass of a continuous object.
5. **Probability distributions:** Integrals are used to find probabilities for continuous random variables using probability density functions.

Would you like further details on any of these applications or the integral setup? Here are a few related questions for further exploration:

1. How do you find the volume of a solid of revolution using integrals?
2. Can integrals be used to calculate the arc length of a curve?
3. How is the centroid of a region calculated using integrals?
4. How are integrals applied in physics, specifically for computing work?
5. Can integrals be used in probability to find expected values?

**Tip:** Always visualize the curves or functions when dealing with integral problems. It helps in setting the correct limits and determining which function goes on top.

Learn how to calculate the area between the curves x = y^2 and x = 1 - y^2 using integral calculus. This problem demonstrates finding points of intersection and setting up integrals to compute the area between two parabolas. Suitable for college-level calculus students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation