Question

In Exercises \(11-22,\) evaluate the iterated integral. $$ \int_{0}^{1} \int_{0}^{x} \sqrt{1-x^{2}} d y d x $$

Short answer

The value of the integral is \(-\frac{1}{3}\).

Step-by-step solution

Integrate Outer Integral

Begin by solving the outermost integral. This is the integral from \(0\) to \(x\) which encompasses the entire problem. Here, because the limits of the inner integral involve \(y\), it is possible to integrate with respect to \(y\) by treating \(x\) as a constant. The integral of \(\sqrt{1-x^{2}}\) with respect to \(y\) is \(y\sqrt{1-x^{2}}\) and evaluating this from \(0\) to \(x\) gives \(x\sqrt{1-x^{2}}\). Thus, this simplifies inner integral to \(x\sqrt{1-x^{2}}\). So, the task now is to evaluate \(\int_{0}^{1} x\sqrt{1-x^{2}} dx\).

Integrate Simplified Problem

The next available step is to calculate the remaining single integral. However, this is non-trivial integral which requires substitution. Let \(u=1-x^{2}\). Then, \(du=-2xdx\) and \(xdx=-\frac{1}{2} du\). Thus, the integral becomes \(\int_{1}^{0} \frac{1}{2}\sqrt{u} du\), but this has limits backwards, so it equals to-\(\int_{0}^{1} \frac{1}{2}\sqrt{u} du=-\frac{1}{2}[\frac{2}{3}u^\frac{3}{2}]\Bigg|_{0}^{1}\).

Compute and Simplify the Result

Solving the integral yields -\(\frac{1}{2}[\frac{2}{3}1^\frac{3}{2}-\frac{2}{3}0^\frac{3}{2}]=-1/3\). The negative comes from the change of the limits of integration when applying the substitution.

Key concepts

Double Integral

The concept of a double integral is at the heart of calculating the volume under a surface in a two-dimensional region. It involves integrating a function of two variables, say, for example, over a rectangular region in the xy-plane. The process can be thought of as adding up infinitesimally small rectangles, each with an area of \(dy \cdot dx\), over the defined limits.

• First, an inner integral is evaluated, holding the outer variable constant.
• Then, the result of the inner integral is used to find the outer integral, effectively reducing a two-dimensional problem to a one-dimensional one.

In the context of the given exercise, the double integral is set up to integrate the function \( \sqrt{1-x^2} \) over a triangle in the xy-plane with vertices at (0,0), (1,0), and (1,1). Such an approach allows for finding the cumulative sum along the y-direction first (from 0 to x, hence the variable upper limit) and then along the x-direction (from 0 to 1).

Integration by Substitution

Integration by substitution is a technique used to simplify an integral by changing variables. It is comparable to the 'u-substitution' used in single-variable calculus. When an integral is difficult to tackle with a standard formula, a substitution can often transform it into an easier form.

• A new variable \(u\) is introduced to replace an expression within the integral.
• The differential \(dx\) is then expressed in terms of \(du\), following the chain rule of differentiation.
• Finally, we replace the integrand and the limits of integration and solve with respect to \(u\).
• Don't forget to change the limits of integration to match the new variable, or alternatively, change back to the original variable after integrating.

In the exercise, the substitution \(u = 1 - x^2\) and \(du = -2xdx\) transforms the integral to one in terms of \(u\), making it easier to evaluate. However, the process in the solution inadvertently reversed the limits of integration, hence the negative sign in front of the integral.

Definite Integral

A definite integral represents the signed area under a curve between two points on the x-axis. In a physical sense, it can represent total accumulation of quantities like area, volume, and mass. The properties of definite integrals include:

• Being evaluated over a closed interval \[a, b\].
• The Fundamental Theorem of Calculus links the definite integral to the antiderivative of the function.
• The result of a definite integral is a number, as opposed to the antiderivative, which is a function.

As applied to our exercise, after using integration by substitution, we compute the definite integral from the updated limits \( \sqrt{u} \) from 0 to 1. The solution correctly shows that after finding the antiderivative, we must substitute the upper and lower limits of the interval into the antiderivative and subtract the latter from the former. The resulting number, -1/3, represents the signed area under the original curve within the triangle mentioned earlier.