Question

In Exercises \(41-44\), set up and evaluate the integrals for finding the area and moments about the \(x\) - and y-axes for the region bounded by the graphs of the equations. (Assume \(\rho=1\).) $$ y=x^{2}-4, y=0 $$

Short answer

The area under the curves is 0 square units and the moments about the x and y axes are also 0.

Step-by-step solution

Establish Limits of Integration

First, identify the intersection points of \(y=x^2-4\) and \(y=0\) to establish the limits of the integrals. To do this, set \(x^2-4\) equal to 0 and solve for \(x\). The solutions \(x=-2\) and \(x=2\) will give the limits of the integrals.

Set Up and Evaluate Area Integral

Next, set up the integral for finding the area. The area \(A\) is defined as the integral of \(y\) dx from \(x=-2\) to \(x=2\): \(A=\int_{-2}^{2} (x^2-4) dx\). Now evaluate this integral. The antiderivative of \(x^2-4\) is \(\frac{1}{3}x^3-4x\), so the integral evaluates to: \(\frac{1}{3}(2)^3 - 4*(2) - (\frac{1}{3}(-2)^3 - 4*(-2)) = 0\). Therefore, the area is 0 square units.

Set Up and Evaluate Moments Integrals

Now we calculate the moments around the x-axis and the y-axis. The moment M_x around the x-axis is defined as the integral of \(y^2\) dx from \(x=-2\) to \(x=2\): \(M_x=\int_{-2}^{2} (x^4 - 8x^2 +16) dx\). Similarly, the moment M_y around the y-axis is \(M_y=\int_{-2}^{2} x(x^2 -4) dx\). The antiderivative of \(x^4 - 8x^2 +16\) is \(\frac{1}{5}x^5 - \frac{8}{3}x^3 + 16x\); and for \(x(x^2-4)\) is \(\frac{1}{4}x^4 - 4x^2\). Solving these integrals give us the following results: \(M_x=0\) and \(M_y=0\). Therefore, the moments about the x and y axes are 0.

Key concepts

Limits of Integration

When dealing with calculus problems, particularly those involving area under curves, it's pivotal to understand limits of integration. These limits define the interval on the x-axis over which we will integrate the function. For example, when asked to find the area of a region bounded by the graphs of the equations y=x^2-4 and y=0, we first need to establish where these two graphs intersect. This occurs at the points where the function equals zero, hence x^2 - 4 = 0. Solving this equation gives us the points x=-2 and x=2, which serve as our limits of integration. These points are crucial as they bound the area on the graph that we are interested in calculating.

Choosing the correct limits of integration ensures that we evaluate the integral over the exact interval we are interested in, which is essential for accurately determining areas and moments in calculus. It's the difference between measuring the entire length of a piece of string versus just a segment of it.

Evaluating Integrals

Once we have established our limits of integration, the next step is evaluating integrals. This process involves finding the antiderivative of the given function and applying the Fundamental Theorem of Calculus. For instance, to find the area under the graph of y=x^2-4 from x=-2 to x=2, we integrate the function with respect to x over this interval. We determine the integral of x^2-4, which gives us the antiderivative \( \frac{1}{3}x^3 - 4x \).

We then evaluate this expression at our limits of integration, which has the effect of 'plugging in' these values into our antiderivative to get the definite integral's value. In our example, it turns out that the area is zero, which might at first seem surprising. However, this happens because the function is symmetrical around the x-axis and the positive area above the axis exactly cancels out the negative area below it. Such exercises reinforce the importance of not only knowing how to integrate a function, but also understanding the geometric interpretation of these integrals.

Moments of Inertia

In calculus, moments of inertia quantify how mass is distributed around an axis. Moments are crucial in engineering and physics for determining the rotational behavior of objects. Suppose we have a lamina, a flat two-dimensional surface, of uniform density and we wish to determine its rotation around the x or y-axis. Moments about the x or y-axis, denoted as M_x and M_y, respectively, are found by integrating specific functions across the region of interest. For a region bounded by y = x^2 -4 and y = 0, the moments are calculated as the integrals of y² and xy respectively, over the limits established by the intersection points.

Moments of inertia are also a measure of an object's resistance to changes in rotation rate. In our example, although the calculation of the moments results in 0, in practice, moments of inertia provide insight into how much energy is needed for an object to achieve a certain rotational speed. This leverage of calculus for practical applications makes it an invaluable tool in the scientific and engineering domains.