To the right is displayed what the solid of revolution would look like if you rotated the displayed area about the x-axis. As an exercise, try to calculate this volume and see how your answer compares to the volume displayed. (Note: For volumes that are irrational, the displayed value only approximates the true exact value.).
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Section11.2Iterated Integrals
Recall that we defined the double integral of a continuous function (f = f(x,y)) over a rectangle (R = [a,b] times [c,d]) as
Solids of revolution are used commonly in engineering and manufacturing. Some examples are axles, funnels, pills, bottles, and pistons, as shown in Figure 7.12. Solids of revolution Figure 7.12 When a region in the plane is revolved about a line, the resulting solid is a solid of revolution, and the line is called the axis of revolution. Answer to: Find the volume of the torus generated by revolving the circle (x - 11)^2 + y^2 = 49 about the y-axis. (Give an exact answer, using pi.
where the different variables and notation are as described in Section 11.1. Thus (iint_R f(x,y) , dA) is a limit of double Riemann sums, but while this definition tells us exactly what a double integral is, it is not very helpful for determining the value of a double integral. Fortunately, there is a way to view a double integral as an iterated integral, which will make computations feasible in many cases.
The viewpoint of an iterated integral is closely connected to an important idea from single-variable calculus. When we studied solids of revolution, such as the one shown in Figure 11.2.1, we saw that in some circumstances we could slice the solid perpendicular to an axis and have each slice be approximately a circular disk. From there, we were able to find the volume of each disk, and then use an integral to add the volumes of the slices. In what follows, we are able to use single integrals to generalize this approach to handle even more general geometric shapes.
Subsection11.2.1Iterated Integrals
The ideas that we explored in Preview Activity 11.2.1 work more generally and lead to the idea of an iterated integral. Let (f) be a continuous function on a rectangular domain (R = [a,b] times [c,d]text{,}) and let
The function (A = A(x)) determines the value of the cross sectional area (by area we mean “signed” area) in the (y) direction for the fixed value of (x) of the solid bounded between the surface defined by (f) and the (xy)-plane.
The value of this cross sectional area is determined by the input (x) in (Atext{.}) Since (A) is a function of (xtext{,}) it follows that we can integrate (A) with respect to (xtext{.}) In doing so, we use a partition of ([a, b]) and make an approximation to the integral given by
where (x_i^*) is any number in the subinterval ([x_{i-1},x_i]text{.}) Each term (A(x_i^*) Delta x) in the sum represents an approximation of a fixed cross sectional slice of the surface in the (y) direction with a fixed width of (Delta x) as illustrated in Figure 11.2.3. We add the signed volumes of these slices as shown in the frames in Figure 11.2.3 to obtain an approximation of the total signed volume.
As we let the number of subintervals in the (x) direction approach infinity, we can see that the Riemann sum (sum_{i=1}^m A(x_i^*) Delta x) approaches a limit and that limit is the sum of signed volumes bounded by the function (f) on (Rtext{.}) Therefore, since (A(x)) is itself determined by an integral, we have
Hence, we can compute the double integral of (f) over (R) by first integrating (f) with respect to (y) on ([c, d]text{,}) then integrating the resulting function of (x) with respect to (x) on ([a, b]text{.}) The nested integral
is called an iterated integral, and we see that each double integral may be represented by two single integrals.
We made a choice to integrate first with respect to (ytext{.}) The same argument shows that we can also find the double integral as an iterated integral integrating with respect to (x) first, or
The fact that integrating in either order results in the same value is known as Fubini's Theorem.
Fubini's theorem enables us to evaluate iterated integrals without resorting to the limit definition. Instead, working with one integral at a time, we can use the Fundamental Theorem of Calculus from single-variable calculus to find the exact value of each integral, starting with the inner integral.
Subsection11.2.2Summary
We can evaluate the double integral (iint_R f(x,y) , dA) over a rectangle (R = [a,b] times [c,d]) as an iterated integral in one of two ways:
-.
(int_a^b left( int_c^d f(x,y) , dy right) , dxtext{,}) or
-.
(int_c^d left( int_a^b f(x,y) , dx right) , dytext{.})
This process works because each inner integral represents a cross-sectional (signed) area and the outer integral then sums all of the cross-sectional (signed) areas. Fubini's Theorem guarantees that the resulting value is the same, regardless of the order in which we integrate.
Exercises11.2.3Exercises
Section11.2Iterated Integrals
Recall that we defined the double integral of a continuous function (f = f(x,y)) over a rectangle (R = [a,b] times [c,d]) as
where the different variables and notation are as described in Section 11.1. Thus (iint_R f(x,y) , dA) is a limit of double Riemann sums, but while this definition tells us exactly what a double integral is, it is not very helpful for determining the value of a double integral. Fortunately, there is a way to view a double integral as an iterated integral, which will make computations feasible in many cases.
The viewpoint of an iterated integral is closely connected to an important idea from single-variable calculus. When we studied solids of revolution, such as the one shown in Figure 11.2.1, we saw that in some circumstances we could slice the solid perpendicular to an axis and have each slice be approximately a circular disk. From there, we were able to find the volume of each disk, and then use an integral to add the volumes of the slices. In what follows, we are able to use single integrals to generalize this approach to handle even more general geometric shapes.
Subsection11.2.1Iterated Integrals
The ideas that we explored in Preview Activity 11.2.1 work more generally and lead to the idea of an iterated integral. Let (f) be a continuous function on a rectangular domain (R = [a,b] times [c,d]text{,}) and let
11.2 Solids Of Revolution Discap Calculus 2nd Edition
The function (A = A(x)) determines the value of the cross sectional area (by area we mean “signed” area) in the (y) direction for the fixed value of (x) of the solid bounded between the surface defined by (f) and the (xy)-plane.
The value of this cross sectional area is determined by the input (x) in (Atext{.}) Since (A) is a function of (xtext{,}) it follows that we can integrate (A) with respect to (xtext{.}) In doing so, we use a partition of ([a, b]) and make an approximation to the integral given by
where (x_i^*) is any number in the subinterval ([x_{i-1},x_i]text{.}) Each term (A(x_i^*) Delta x) in the sum represents an approximation of a fixed cross sectional slice of the surface in the (y) direction with a fixed width of (Delta x) as illustrated in Figure 11.2.3. We add the signed volumes of these slices as shown in the frames in Figure 11.2.3 to obtain an approximation of the total signed volume.
As we let the number of subintervals in the (x) direction approach infinity, we can see that the Riemann sum (sum_{i=1}^m A(x_i^*) Delta x) approaches a limit and that limit is the sum of signed volumes bounded by the function (f) on (Rtext{.}) Therefore, since (A(x)) is itself determined by an integral, we have
Hence, we can compute the double integral of (f) over (R) by first integrating (f) with respect to (y) on ([c, d]text{,}) then integrating the resulting function of (x) with respect to (x) on ([a, b]text{.}) The nested integral
11.2 Solids Of Revolution Discap Calculus 14th Edition
is called an iterated integral, and we see that each double integral may be represented by two single integrals.
We made a choice to integrate first with respect to (ytext{.}) The same argument shows that we can also find the double integral as an iterated integral integrating with respect to (x) first, or
The fact that integrating in either order results in the same value is known as Fubini's Theorem.
11.2 Solids Of Revolution Discap Calculus Solver
Fubini's theorem enables us to evaluate iterated integrals without resorting to the limit definition. Instead, working with one integral at a time, we can use the Fundamental Theorem of Calculus from single-variable calculus to find the exact value of each integral, starting with the inner integral.
Subsection11.2.2Summary
11.2 Solids Of Revolution Discap Calculus Calculator
We can evaluate the double integral (iint_R f(x,y) , dA) over a rectangle (R = [a,b] times [c,d]) as an iterated integral in one of two ways:
-.
(int_a^b left( int_c^d f(x,y) , dy right) , dxtext{,}) or
-.
(int_c^d left( int_a^b f(x,y) , dx right) , dytext{.})
This process works because each inner integral represents a cross-sectional (signed) area and the outer integral then sums all of the cross-sectional (signed) areas. Fubini's Theorem guarantees that the resulting value is the same, regardless of the order in which we integrate.
Exercises11.2.3Exercises