## Calculus AB and Calculus BC

## CHAPTER 7 Applications of Integration to Geometry

**Concepts and Skills**

In this chapter, we will review using definite integrals to find areas and volumes; specifically

• area under a curve,

• area between two curves,

• volumes of solids with known cross sections,

• and volumes of solids of revolution (using disks and washers).

We’ll also review related BC topics, including

• arc length;

• arc lengths, areas, and volumes involving parametrically defined functions;

• and area and arc length for polar curves.

Also for BC Calculus students, we’ll review the topic of improper integrals, including

• recognizing when an integral is improper

• and techniques for determining whether an improper integral converges or diverges.

**A. AREA**

To find an area, we

(1) draw a sketch of the given region and of a typical element;

(2) write the expression for the area of a typical rectangle; and

(3) set up the definite integral that is the limit of the Riemann sum of *n* areas as *n* → ∞.

**FIGURE N7–1**

If *f* (*x*) is nonnegative on [*a*,*b*], as in Figure N7–1, then *f* (*x _{k}*) Δ

*x*can be regarded as the area of a typical approximating rectangle, and the area bounded by the

*x*-axis, the curve, and the vertical lines

*x*=

*a*and

*x*=

*b*is given exactly by

See Questions 1,5, and 10 in the Practice Exercises at the end of this chapter.

If *f* (*x*) changes sign on the interval (Figure N7–2), we find the values of *x* for which *f* (*x*) = 0 and note where the function is positive, where it is negative. The total area bounded by the *x*-axis, the curve, *x* = *a,* and *x* = *b* is here given exactly by

where we have taken into account that *f* (*x _{k}*) Δ

*x*is a negative number if

*c*<

*x*<

*d*.

**FIGURE N7–2**

See Question 11 in the Practice Exercises.

If *x* is given as a function of *y*, say *x* = *g*(*y*), then (Figure N7–3) the subdivisions are made along the *y*-axis, and the area bounded by the *y*-axis, the curve, and the horizontal lines *y* = *a* and *y* = *b* is given exactly by

See Questions 3 and 13 in the Practice Exercises.

**FIGURE N7–3**

**A1. Area Between Curves.**

To find the area between curves (Figure N7–4), we first find where they intersect and then write the area of a typical element for each region between the points of intersection. For the total area bounded by the curves *y* = *f* (*x*) and *y* = *g*(*x*) between *x* = *a* and *x* = *e*, we see that, if they intersect at [*c*,*d*], the total area is given exactly by

See Questions 4, 6, 7, and 9 in the Practice Exercises.

**FIGURE N7–4**

**A2. Using Symmetry.**

Frequently we seek the area of a region that is symmetric to the *x*- or *y*-axis (or both) or to the origin. In such cases it is almost always simpler to make use of this symmetry when integrating. For example:

• The area bounded by the *x*-axis and this arch of the cosine curve is symmetric to the *y*-axis; hence it is twice the area of the region to the right of the *y*-axis.

• The area bounded by the parabola and the line is symmetric to the *x*-axis; hence it is twice the area of the region above the *x*-axis.

• The ellipse is symmetric to both axes; hence the area inside the ellipse is four times the area in the first quadrant.

**Evaluating** **Using a Graphing Calculator**

The calculator is especially useful in evaluating definite integrals when the *x*-intercepts are not easily determined otherwise or when an explicit antiderivative of *f* is not obvious (or does not exist).

**EXAMPLE 1**

Evaluate

**SOLUTION:** The integrand *f* (*x*) = *e*^{−}^{x}^{2} has no easy antiderivative. The calculator estimates the value of the integral to be 0.747 to three decimal places.

**EXAMPLE 2**

In Figure N7–5, find the area under *f* (*x*) = −*x*^{4} + *x*^{2} + *x* + 10 and above the *x*-axis.

**FIGURE N7–5**

**SOLUTION:** To get an accurate answer for the area use the calculator to find the two intercepts, storing them as *P* and *Q*, and then evaluate the integral:

which is accurate to three decimal places.

**Region Bounded by a Parametric Curve**

If *x* and *y* are given parametrically, say by *x* = *f* (θ), *y* = *g*(θ), then to evaluate we express *y*, *dx*, and the limits *a* and *b* in terms of θ and *d*θ, then integrate. Remember that we define *dx* to be *x* *′*(θ) *d*θ, or *f* *′*(θ) *d*θ.

See Questions 14, 15, and 44 in the Practice Exercises.

**BC ONLY**

**Region Bounded by Polar Curve**

**FIGURE N7–6**

To find the area *A* bounded by the polar curve *r* = *f* (θ) and the rays θ = *α* and θ = β (see Figure N7–6), we divide the region into *n* sectors like the one shown. If we think of that element of area, Δ*A*, as a circular sector with radius *r* and central angle Δθ, its area is given by

Summing the areas of all such sectors yields the area of the entire region:

The expression above is a Riemann sum, equivalent to this definite integral:

We have assumed above that *f* (θ) 0 on [α, β]. We must be careful in determining the limits α and β in (2); often it helps to think of the required area as that “swept out” (or generated) as the radius vector (from the pole) rotates from θ = α to θ = β. It is also useful to exploit symmetry of the curve wherever possible.

The relations between rectangular and polar coordinates, some common polar equations, and graphs of polar curves are given in the Appendix.

**BC ONLY**

**EXAMPLE 3**

Find the area inside both the circle *r* = 3 sin θ and the cardioid *r* = 1 + sin θ.

**SOLUTION:** Choosing an appropriate window, graph the curves on your calculator.

See Figure N7–7, where one half of the required area is shaded. Since 3 sin θ = 1 + sin θ when we see that the desired area is twice the sum of two parts: the area of the circle swept out by θ as it varies from 0 to plus the area of the cardioid swept out by a radius vector as θ varies from Consequently

**FIGURE N7–7**

See also Questions 46 and 47 in the Practice Exercises.

**BC ONLY**

**EXAMPLE 4**

Find the area enclosed by the cardioid *r* = 2(1 + cos θ).

**SOLUTION:** We graphed the cardioid on our calculator, using polar mode, in the window [−2,5] × [−3,3] with θ in [0,2π].

**FIGURE N7–8**

Using the symmetry of the curve with respect to the polar axis we write