Question

Find the volume of the following solid regions. The solid above the parabolic region \(R=\\{(x, y): 0 \leq x \leq 1\), \(\left.0 \leq y \leq 1-x^{2}\right\\}\) and between the planes \(z=1\) and \(z=2-y\)

Short answer

Answer: \(\frac{1}{2}\)

Step-by-step solution

Set up the double integral for the volume

Set up a double integral for the volume of the solid region described by \(0 \leq x \leq 1\), \(0 \leq y \leq 1-x^2\), and the difference in z between \(z=1\) and \(z=2-y\). The double integral will be of the form: $$V = \int\int_R (2-y-1) \, dA$$

Decide on the integration order

Since the region R is given in terms of x and y, it's easier to integrate with respect to y first, then x. This gives us: $$ V = \int_{0}^{1} \int_{0}^{1-x^2} (1-y) \, dy \, dx $$

Evaluate the inner integral

Evaluate the inner integral, which is with respect to y: $$ \int_{0}^{1-x^2} (1-y) \,dy = \left[y - \frac{1}{2}y^2 \right]_0^{1-x^2} $$ Substitute the limits into the brackets and simplify the expression: $$ \left[(1-x^2) - \frac{1}{2}(1-x^2)^2 \right] - \left[0 - \frac{1}{2}(0)^2 \right] = (1-x^2) - \frac{1}{2}(1-2x^2+x^4) $$

Evaluate the outer integral

Now, evaluate the outer integral, which is with respect to x: $$ V = \int_{0}^{1} \left((1-x^2) - \frac{1}{2}(1-2x^2+x^4)\right) \, dx $$ Integration of each term will give: $$ V = \left[x - \frac{1}{3}x^3 - \frac{1}{2}x + x^3 - \frac{1}{10}x^5 \right]_0^1 $$

Calculate the volume

Finally, substitute the limit x=1 into the expression and find the volume of the solid: $$ V = \left[(1) - \frac{1}{3}(1)^3 - \frac{1}{2}(1) + (1)^3 - \frac{1}{10}(1)^5 \right] - \left[0 - \frac{1}{3}(0)^3 - \frac{1}{2}(0) + (0)^3 - \frac{1}{10}(0)^5 \right] = \frac{1}{2} $$ The volume of the given solid region is \(\frac{1}{2}\).

Key concepts

Volume of Solid Regions

Understanding the volume of solid regions is a key step in grasping the application of integrals to real-world problems. Volume is essentially the amount of space that a 3-dimensional shape occupies. Mathematicians and engineers often use double integrals to calculate the volume of complex shapes, such as the solid above the parabolic region in the exercise.

Consider a solid object that is situated above a certain region in the xy-plane and is bounded between two parallel planes along the z-axis. The volume of such a solid can be computed by taking the integral of the 'height' of the solid (the z-coordinate) over the area of the region in the xy-plane. The 'height' is actually the difference between the upper and lower bounds of the solid along the z-axis. The integral effectively adds up all the infinitesimally small volumes (dV) across the entire region to find the total volume.

Multiple Integration

Multiple integration refers to the process of integrating a function of several variables over a region in n-dimensional space. In the case of finding the volume of a solid region, we often use double integrals, as they allow us to integrate over a two-dimensional region.

When setting up a double integral, the order of integration (which variable to integrate first) is essential. The choice can greatly simplify the calculation. Generally, choosing the order depends on the limits of integration and the function being integrated. For instance, if the region's x-boundaries are simple, but the y-boundaries are complicated functions of x, it may be easier to integrate with respect to y first. Multiple integration enables us to solve problems that involve more than one variable, leading to a wide array of applications in physics, engineering, and probability.

Parabolic Region

A parabolic region is an area in the xy-plane bounded by a parabola and straight lines, which in some cases represents the projection of a three-dimensional object onto a plane. The exercise defines a parabolic region bounded by the parabola given by the equation \(y = 1-x^2\) and the lines \(y = 0\) and \(x = 1\).

Geometrically, this region has curved sides because of the parabola. In the context of integrals, these curved boundaries create varying limits for the integration. For such regions, one of the key steps in setting up an integral for the volume calculation is correctly identifying the limits for integration, which must encapsulate the whole area under the curve between the specified lines.

Integration Limits

Integration limits define the range over which the integration is performed. In single-variable calculus, the limits are simply two values between which the integral is calculated. However, when dealing with multiple integrals, such as double integrals, the limits become more complex as they can be functions that depend on the other variables.

In the given exercise, the limits for y are dependent on x, ranging from 0 to \(1-x^2\), which signifies that for each value of x, y varies between the parabola and the x-axis. The x-limits are constant, from 0 to 1, covering the full extent of the region in the x direction. Correctly calculating these limits is critical and often requires a deep understanding of the problem's geometry.