Question

Sketch the region bounded by the graphs of the given equations, show a typical slice, approximate its area, set up an integral, and calculate the area of the region. Make an estimate of the area to confirm your answer. \(4 y^{2}-2 x=0,4 y^{2}+4 x-12=0\)

Short answer

The area of the region is 4 square units.

Step-by-step solution

Simplify and Rearrange Equations

First, simplify and rearrange the given equations to make them more manageable and identify intersections.For the first equation, \(4y^2 - 2x = 0\), solve for \(x\):\[ x = 2y^2 \]For the second equation, \(4y^2 + 4x - 12 = 0\), solve for \(x\):\[ x = 3 - y^2 \]These are the functions: \(x = 2y^2\) and \(x = 3 - y^2\).

Determine Points of Intersection

To find the points where these curves intersect, set the equations equal to each other:\[ 2y^2 = 3 - y^2 \]\[ 3y^2 = 3 \]\[ y^2 = 1 \]Thus, \(y = 1\) or \(y = -1\) are the intersections. Substitute these back into either equation to solve for \(x\):For \(y = 1\), \(x = 2 \times (1)^2 = 2\).For \(y = -1\), \(x = 2 \times (-1)^2 = 2\).Points of intersection are \((2, 1)\) and \((2, -1)\).

Sketch and Identify Typical Slice

Draw the graphs \(x = 2y^2\) and \(x = 3 - y^2\) on the coordinate plane. The region of interest is between \((2, -1)\) and \((2, 1)\). Sketch a typical slice, which is a vertical strip spanning from \(x = 2y^2\) to \(x = 3 - y^2\) at any \(y\).

Set up the Integral for Area

The area of the region can be calculated by integrating the width of the vertical slice from \(y = -1\) to \(y = 1\).The expression for the width of a slice at a particular \(y\) is:\[ (3 - y^2) - (2y^2) = 3 - 3y^2 \]Set up the integral for the area:\[ A = \int_{-1}^{1} (3 - 3y^2) \, dy \]

Compute the Integral

Calculate the integral to find the area:\[ A = \int_{-1}^{1} (3 - 3y^2) \, dy = \left[ 3y - y^3 \right]_{-1}^{1} \]Evaluate at the boundaries:\[ A = \left( 3 \times 1 - 1^3 \right) - \left( 3 \times (-1) - (-1)^3 \right) \]\[ A = (3 - 1) - (-3 + 1) = 2 - (-2) = 4 \]

Estimate the Area

Using the symmetrical layout of the region in the sketch, divide it into simple geometric shapes. Observe that each half (top and bottom) can be approximated visually to have areas around 2 units. Total estimated area is close to 4 square units, matching our calculus result.

Key concepts

Integral Calculus

Integral calculus is a fundamental part of calculus, which is focused on accumulation and the area beneath curves. It involves finding antiderivatives, which are essentially functions that can be used to calculate the net area under a curve over a given interval. One of the core objectives in integral calculus is to determine the exact area, even when the shape is irregular and not easily measurable with standard geometric formulas.

To solve problems using integral calculus, one must set up an integral, which is a mathematical representation of the accumulated area. This involves integrating a function over a defined interval. For instance, in the exercise, the integral \[ \int_{-1}^{1} (3 - 3y^2) \, dy \] was outlined. This integral sums up the differences in lengths of slices, calculated from the two curves bounding the area. Hence, integral calculus systematically allows us to find the total area enclosed by curves on a graph.

Area Under Curves

Calculating the area under curves is a quintessential skill in calculus, especially when dealing with irregular shapes on a graph. To find the area under a curve, you integrate the function that represents the curve over a specific interval.

In our exercise, we identified the two curves represented by the equations \( x = 2y^2 \) and \( x = 3 - y^2 \). The area that is sought lies between these two curves from \( y = -1 \) to \( y = 1 \). By calculating the area under each curve separately and finding the difference, integral calculus allows us to precisely find the area enclosed between them. This systematic process helps illustrate the concept that the integral can be viewed as the 'sum' of infinitely small slices that fill the area between curves.

Intersection Points

Intersection points are where two or more curves cross each other. Finding these points is essential because they often define the bounds of areas of interest. In many calculus problems, the intersection points give the limits for the integration.

In the case of the original exercise, we found intersection points by equating the given functions: \(2y^2 = 3 - y^2\). Solving this equation enabled us to find \(y = 1\) and \(y = -1\), which were crucial to determining where the curves intersected. These specific intersection points translated to the \(x\) value of 2 for both \(y = 1\) and \(y = -1\), providing the vertical lines marking the area we needed to calculate. Accurate determination of these intersection points ensures that we are evaluating the area over the correct domain.

Graph Sketching

Sketching a graph is a visual aid that helps in understanding the behavior of functions and identifying key features such as intersection points, symmetry, and bounded regions. A sketch provides a concrete representation of abstract equations, making it easier to conceptualize problems.

In our problem, sketching the graphs of the equations \( x = 2y^2 \) and \( x = 3 - y^2 \) was critical. It helped us visualize how these two curves interacted, specifically showing the region between \(2, -1\) and \(2, 1\). A carefully drawn graph reveals the region's symmetry and boundaries, serving as a double-check for calculated areas. Furthermore, when complemented with mathematical calculations, such sketches enhance comprehension and highlight practical application of integral calculus in solving real-world problems.