Sketch the region enclosed by y = 3x and y = 9x^2 . Find the area of the region.

We are tasked with sketching the region enclosed by the curves \( y = 3x \) and \( y = 9x^2 \), and then finding the area of that region.

### Step 1: Finding the points of intersection
To find the points where these two curves intersect, we set the equations equal to each other:

\[
3x = 9x^2
\]

Simplifying this equation:

\[
x(3 - 9x) = 0
\]

Thus, the solutions are:

\[
x = 0 \quad \text{or} \quad x = \frac{1}{3}
\]

So, the curves intersect at \( x = 0 \) and \( x = \frac{1}{3} \).

### Step 2: Determining the enclosed region
We now know that the region of interest is between \( x = 0 \) and \( x = \frac{1}{3} \). The curve \( y = 9x^2 \) lies below \( y = 3x \) in this interval, as we can check by comparing the values of the functions for points within the interval \( [0, \frac{1}{3}] \).

### Step 3: Setting up the integral for the area
The area between two curves is given by the integral of the difference between the top function and the bottom function over the interval of interest:

\[
\text{Area} = \int_{0}^{\frac{1}{3}} (3x - 9x^2) \, dx
\]

### Step 4: Solving the integral

We now compute the integral:

\[
\int (3x - 9x^2) \, dx = \frac{3x^2}{2} - 3x^3
\]

Evaluating this from \( 0 \) to \( \frac{1}{3} \):

\[
\left[ \frac{3x^2}{2} - 3x^3 \right]_{0}^{\frac{1}{3}} = \left( \frac{3(\frac{1}{3})^2}{2} - 3\left(\frac{1}{3}\right)^3 \right) - \left( \frac{3(0)^2}{2} - 3(0)^3 \right)
\]

Simplifying the expression:

\[
= \left( \frac{3 \cdot \frac{1}{9}}{2} - 3 \cdot \frac{1}{27} \right) = \left( \frac{1}{6} - \frac{1}{9} \right)
\]

Now, finding a common denominator:

\[
= \frac{3}{18} - \frac{2}{18} = \frac{1}{18}
\]

Thus, the area of the enclosed region is \( \frac{1}{18} \) square units.

### Final Answer:
The area of the region enclosed by the curves \( y = 3x \) and \( y = 9x^2 \) is \( \frac{1}{18} \).

Would you like me to sketch the region for visualization?

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#### 5 Related Questions:
1. What is the geometric meaning of the integral used to find the area between two curves?
2. How would the area change if the parabola equation was \( y = 9x^2 + 1 \)?
3. Can we find the volume of a solid generated by rotating the region around the x-axis?
4. What happens to the intersection points if we change the slope of the line to \( y = 4x \)?
5. How would you calculate the centroid of the enclosed region?

#### Tip:
When comparing two curves, always make sure to identify which one lies above the other within the interval to correctly set up the area integral.

Sketch the region enclosed by y = 3x and y = 9x^2. Find the area of the region.

Explore how to find the area enclosed by the curves y = 3x and y = 9x^2. This calculus problem demonstrates the application of integrals to compute areas between curves, using the intersection points and integral setup. Ideal for Grades 11-12 and college-level calculus students.

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