Find the areas of the regions enclosed by the curves and lines in Exercises 15–26.
x = 2y², x = 0, y = 3

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First, identify the region enclosed by the curves and lines given: $x = 2y^{2}$, $x = 0$, and $y = 3$. Note that $x = 0$ is the y-axis, and $y = 3$ is a horizontal line.

Determine the limits for $y$. Since $x = 0$ and $y = 3$ are boundaries, and $x = 2y^{2}$ opens to the right, the region is bounded between $y = 0$ (implicitly, since $x = 0$ intersects $y=0$) and $y = 3$.

Set up the integral for the area. The area between the curves can be found by integrating the horizontal distance between the curves with respect to $y$. The right boundary is $x = 2y^{2}$ and the left boundary is $x = 0$, so the width at each $y$ is $2y^{2} - 0 = 2y^{2}$.

Write the integral expression for the area as $A = \int_{0}^{3} (2y^{2}) \, dy$.

Evaluate the integral by integrating \$2y^{2}$ with respect to $y$ over the interval $[0,3]$ to find the enclosed area.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Area Between Curves

The area between curves is found by integrating the difference of the functions that define the boundaries over the interval of interest. When curves are given implicitly or in terms of y, the integration may be with respect to y or x, depending on the orientation of the region.

Setting Up the Integral with Respect to y

Since the curve is given as x = 2y², and the boundaries include x = 0 and y = 3, it is natural to integrate with respect to y. The limits of integration are from y = 0 to y = 3, and the integrand is the horizontal distance between the curves, here x = 2y² minus x = 0.

Evaluating Definite Integrals

After setting up the integral, evaluating it involves applying the power rule for integration and substituting the upper and lower limits. This process yields the exact area of the enclosed region, providing a numerical value representing the size of the region.

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