Problem 73 - University Calculus: Early Transcendentals

Question

Find the areas of the regions enclosed by the lines and curves in Exercises \(73-80\). $$x=2 y^{2}, \quad x=0, \quad \text { and } \quad y=3$$

Step-by-Step Solution

Answer

The area is 18 square units.

Step 1: Understand the Problem

We need to find the region bounded by the parabola \(x = 2y^2\), the line \(x = 0\) (the y-axis), and \(y = 3\). The area is enclosed between these curves and lines in the xy-plane.

Step 2: Identify Intersection Points

Since \(x = 0\) is the y-axis, the intersections with \(x = 2y^2\) occur at \(x = 0\). Plugging \(x = 0\) into the parabola equation gives \(0 = 2y^2\), which means \(y = 0\). Therefore, one point of intersection is at the origin \((0, 0)\). Another boundary given is \(y = 3\), which represents a horizontal line, where it intersects the parabola at \(x=2 \times 3^2 = 18\). Thus, the intersections are at \((0, 0)\) and \((18, 3)\).

Step 3: Set Up the Integral

The area is bounded by the y-axis (\(x = 0\)), the parabola (\(x = 2y^2\)), and the line \(y = 3\). We can express the area as an integral with respect to \(y\) from \(y = 0\) to \(y = 3\), integrating the function \(x = 2y^2\) minus zero (the y-axis): \[ A = \int_{0}^{3} (2y^2)\, dy \]

Step 4: Evaluate the Integral

Compute the integral:\[ A = \int_{0}^{3} 2y^2 \, dy = 2 \left[ \frac{y^3}{3} \right]_{0}^{3} = 2 \left( \frac{3^3}{3} - \frac{0^3}{3} \right) \] Simplify:\[ A = 2 \left( \frac{27}{3} \right) = 2 \times 9 = 18 \]

Step 5: Conclusion

The area of the region enclosed by the parabola, y-axis, and the line \(y=3\) is \(18\) square units.

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, especially when dealing with area calculations. When we talk about definite integrals, we're referring to the accumulation of quantities—here, specifically, the area under a curve, between two limits of integration. In our exercise, we're considering the values from

• the point where the curve starts,
• to the point where it stops,

which are from

• the lower limit at \(y = 0\)
• to the upper limit at \(y = 3\).

The integral of the function \(2y^2\) from 0 to 3 helps us establish how the area is formed across these limits. By plugging in the limits of the integral into the antiderivative, we simplify to find the net area under the curve.
This method offers a precise measurement of space that curves and lines demarcate, making definite integrals a robust tool for examining geometric properties in calculus.

Area Between Curves

Finding the area between curves involves measuring the space enclosed by different functions. In simpler words, we subtract one function from another within the given limits. In our case, the focus is on the region bounded by the parabola \(x = 2y^2\), the line \(x = 0\) (the y-axis), and \(y = 3\).

• The equation \(x = 2y^2\) defines the arc of the parabola, while \(x = 0\) acts as the baseline, or y-axis.
• By evaluating the integral of \(x = 2y^2\) with respect to \(y\) from \(0\) to \(3\),

we compute the precise area enclosed between these curves and lines.
This calculation is essential for understanding how both simple and complex boundaries form enclosed shapes whose areas we can determine through integration.

Parabolic Region

A parabolic region refers to the area enclosed by or around a parabola, which is a type of conic section that opens upward or downward in this context. The equation \(x = 2y^2\) describes a parabola that opens to the right on the xy-plane.
To better understand, think about how this curve stretches over the plane, forming a U-like shape from the origin \((0,0)\) to \((18,3)\) in our context. The symmetry and simplicity of parabolas make these regions easy to manage and compute within calculus.

• This calculation utilizes definite integrals because we are only interested in the shape's finite portion—bounded by the y-axis \(x=0\) and the line \(y=3\).
• By setting these boundaries, we assess and quantify the entire parabolic region by treating the function \(x = 2y^2\) as a tool for finding this area.

Intersection Points

Intersection points are where different curves or lines meet each other in the xy-plane. These points are crucial for setting boundaries in area calculations and understanding where the enclosed region begins and ends.

• For our exercise, we find the intersection points by setting the equations equal and solving for one variable.
• When \(x = 0\), we directly place this value in the parabola's equation, leading to the first intersection at \((0,0)\).
• Another given condition, \(y=3\), intersects the curve at \(x = 18\).

These intersections at

• \((0,0)\)
• \((18,3)\),

help complete the region whose area we are calculating. They are essential delimiters for the integral that leads us to the solution of the area.