 # FURTHER APPLICATIONS OF INTEGRATION

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FURTHER APPLICATIONS OF INTEGRATION
9 FURTHER APPLICATIONS OF INTEGRATION

8.2 Area of a Surface of Revolution
FURTHER APPLICATIONS OF INTEGRATION 8.2 Area of a Surface of Revolution In this section, we will learn about: The area of a surface curved out by a revolving arc.

SURFACE OF REVOLUTION A surface of revolution is formed when a curve is rotated about a line. Such a surface is the lateral boundary of a solid of revolution of the type discussed in Sections 6.2 and 6.3

AREA OF A SURFACE OF REVOLUTION
We want to define the area of a surface of revolution in such a way that it corresponds to our intuition. If the surface area is A, we can imagine that painting the surface would require the same amount of paint as does a flat region with area A .

AREA OF A SURFACE OF REVOLUTION

CIRCULAR CYLINDERS The lateral surface area of a circular cylinder with radius r and height h is taken to be: A = 2πrh We can imagine cutting the cylinder and unrolling it to obtain a rectangle with dimensions of 2πrh and h.

CIRCULAR CONES We can take a circular cone with base radius r and slant height l, cut it along the dashed line as shown, and flatten it to form a sector of a circle with radius and central angle θ = 2πr/l.

CIRCULAR CONES We know that, in general, the area of a sector of a circle with radius l and angle θ is ½ l2 θ.

CIRCULAR CONES So, the area is: Thus, we define the lateral surface area of a cone to be A = πrl.

AREA OF A SURFACE OF REVOLUTION
What about more complicated surfaces of revolution?

AREA OF A SURFACE OF REVOLUTION
If we follow the strategy we used with arc length, we can approximate the original curve by a polygon. When this is rotated about an axis, it creates a simpler surface whose surface area approximates the actual surface area. By taking a limit, we can determine the exact surface area.

BANDS Then, the approximating surface consists of a number of bands—each formed by rotating a line segment about an axis.

BANDS To find the surface area, each of these bands can be considered a portion of a circular cone.

BANDS Equation 1 The area of the band (or frustum of a cone) with slant height l and upper and lower radii r1 and r2 is found by subtracting the areas of two cones:

From similar triangles, we have:
BANDS From similar triangles, we have: This gives:

Putting this in Equation 1, we get
BANDS Formula 2 Putting this in Equation 1, we get or where r = ½(r1 + r2) is the average radius of the band.

AREA OF A SURFACE OF REVOLUTION
Now, we apply this formula to our strategy.

Consider the surface shown here.
SURFACE AREA Consider the surface shown here. It is obtained by rotating the curve y = f(x), a ≤ x ≤ b, about the x-axis, where f is positive and has a continuous derivative.

SURFACE AREA To define its surface area, we divide the interval [a, b] into n subintervals with endpoints x0, x1, , xn and equal width Δx, as we did in determining arc length.

If yi = f(xi), then the point Pi(xi, yi) lies on the curve.
SURFACE AREA If yi = f(xi), then the point Pi(xi, yi) lies on the curve. The part of the surface between xi–1 and xi is approximated by taking the line segment Pi–1 Pi and rotating it about the x-axis.

SURFACE AREA The result is a band with slant height l = | Pi–1Pi | and average radius r = ½(yi–1 + yi). So, by Formula 2, its surface area is:

As in the proof of Theorem 2 in Section 8.1, we have
SURFACE AREA As in the proof of Theorem 2 in Section 8.1, we have where xi* is some number in [xi–1, xi].

SURFACE AREA When Δx is small, we have yi = f(xi) ≈ f(xi*) and yi–1 = f(xi–1) ≈ f(xi*), since f is continuous. Therefore,

SURFACE AREA Formula 3 Thus, an approximation to what we think of as the area of the complete surface of revolution is:

The approximation appears to become better as n → ∞.
SURFACE AREA The approximation appears to become better as n → ∞.

Then, recognizing Formula 3 as a Riemann sum for the function we have:
SURFACE AREA Then, recognizing Formula 3 as a Riemann sum for the function we have:

SURFACE AREA—DEFINITION
Formula 4 Thus, in the case where f is positive and has a continuous derivative, we define the surface area of the surface obtained by rotating the curve y = f(x), a ≤ x≤ b, about the x-axis as:

With the Leibniz notation for derivatives, this formula becomes:
SURFACE AREA Formula 5 With the Leibniz notation for derivatives, this formula becomes:

SURFACE AREA Formula 6 If the curve is described as x = g(y), c ≤ y ≤ d, then the formula for surface area becomes:

SURFACE AREA Formula 7 Then, both Formulas 5 and 6 can be summarized symbolically—using the notation for arc length given in Section 8.1—as:

For rotation about the y-axis, the formula becomes:
SURFACE AREA Formula 8 For rotation about the y-axis, the formula becomes: Here, as before, we can use either or

SURFACE AREA—FORMULAS
You can remember these formulas in the following ways.

SURFACE AREA—FORMULAS
Think of 2πy as the circumference of a circle traced out by the point (x, y) on the curve as it is rotated about the x-axis.

SURFACE AREA—FORMULAS
Think of 2πx s the circumference of a circle traced out by the point (x, y) on the curve as it is rotated about the y-axis.

The curve , –1 ≤ x ≤ 1, is an arc of the circle x2 + y2 = 4 .
SURFACE AREA Example 1 The curve , –1 ≤ x ≤ 1, is an arc of the circle x2 + y2 = 4 . Find the area of the surface obtained by rotating this arc about the x-axis. The surface is a portion of a sphere of radius 2.

SURFACE AREA Example 1 We have:

So, by Formula 5, the surface area is:
Example 1 So, by Formula 5, the surface area is:

Find the area of the resulting surface.
SURFACE AREA Example 2 The arc of the parabola y = x2 from (1, 1) to (2, 4) is rotated about the y-axis. Find the area of the resulting surface.

Using y = x2 and dy/dx = 2x, from Formula 8, we have:
SURFACE AREA E. g. 2—Solution 1 Using y = x2 and dy/dx = 2x, from Formula 8, we have:

Substituting u = 1 + 4x2, we have du = 8x dx.
SURFACE AREA E. g. 2—Solution 1 Substituting u = 1 + 4x2, we have du = 8x dx. Remembering to change the limits of integration, we have:

we have the following solution.
SURFACE AREA E. g. 2—Solution 2 Using x = and dx/dy = , we have the following solution.

SURFACE AREA E. g. 2—Solution 2

SURFACE AREA Example 3 Find the area of the surface generated by rotating the curve y = ex, 0 ≤ x ≤ 1, about the x-axis.

Using Formula 5 with y = ex and dy/dx = ex, we have:
SURFACE AREA Example 3 Using Formula 5 with y = ex and dy/dx = ex, we have:

SURFACE AREA Example 3

Since tan α = e , we have: sec2α = 1 + tan α = 1 + e2 Thus,
SURFACE AREA Example 3 Since tan α = e , we have: sec2α = 1 + tan α = 1 + e2 Thus,