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3. Use the method of Exercise 2 to find the formula for the volume of a right circular cone of height h whose base is a circle of radius R [Figure 17(B)]. (A) 10 R (B) FIGURE 17 Right circular cones. h

3. Use the method of Exercise 2 to find the formula for the volume of a right circular cone of height h whose base is a circle of radius R [Figure 17(B)]. (A) 10 R (B) FIGURE 17 Right circular cones. h

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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2. Let V be the volume of a right circular cone of height 10 whose base is a circle of radius 4 [Figure 17(A)]. (a) Use similar triangles to find the area of a horizontal cross section at a height y. (b) Calculate V by integrating the cross-sectional area. 3. Use the method of Exercise 2 to find the formula for the volume of a right circular cone of height h whose base is a circle of radius R [Figure 17(B)]. (A) - 10 (B) FIGURE 17 Right circular cones.
Transcribed Image Text:2. Let V be the volume of a right circular cone of height 10 whose base is a circle of radius 4 [Figure 17(A)]. (a) Use similar triangles to find the area of a horizontal cross section at a height y. (b) Calculate V by integrating the cross-sectional area. 3. Use the method of Exercise 2 to find the formula for the volume of a right circular cone of height h whose base is a circle of radius R [Figure 17(B)]. (A) - 10 (B) FIGURE 17 Right circular cones.
y A 4 DF FIGURE 3 A horizontal cross section of the pyramid is a square. y=4-x² Cross section is a semicircle of radius √4-y. FIGURE 4 -R B FIGURE 5 12-y Length √4 R 12 r= √R² ² ■ EXAMPLE 1 Volume of a Pyramid Calculate the volume V of a pyramid of height 12 m whose base is a square of side 4 m. Solution To use Eq. (1), we need a formula for the horizontal cross section A(y). Step 1. Find a formula for A(y). Figure 3 shows that the horizontal cross section at height y is a square. To find the sides of this square, apply the law of similar triangles to AABC and to the triangle of height 12-y whose base of lengths lies on the cross section: We find that s = = Step 2. Compute V 12 ² = 6" V Base 2 Height 12 12- y = V = A(y) dy = (12-y), and therefore, A(y) = s² = (12- y)². as the integral of A (y). = 6² 2(12- y) = 6s 12 12 [²²/1 (12- y)² dy = - 2/7 (12 - y) ³¹/1² This agrees with the result obtained using the formula V = Ah for the volume of a pyramid of base A and height h, since Ah = }(4²)(12) = 64. ■ ■ EXAMPLE 2 Compute the volume V of the solid in Figure 4, whose base is the region between the inverted parabola y = 4 - x² and the x-axis, and whose vertical cross sections perpendicular to the y-axis are semicircles. Solution To find a formula for the area A (y) of the cross section, observe that y = 4 - x² can be written x = ±√4-y. We see in Figure 4 that the cross section at y is a semicircle of radius r = √√4-y. This semicircle has area A(y) = r² = (4- y). Therefore, = 64 m³ A(y) dy = =-6²4- 2 π (4-y)dy = (4y - ₁²) = 4x 2 ■ In the next example, we compute volume using vertical rather than horizontal cross sections. This leads to an integral with respect to x rather than y. ■ EXAMPLE 3 Volume of a Sphere: Vertical Cross Sections Compute the volume of a sphere of radius R. Solution As we see in Figure 5, the vertical cross section of the sphere at x is a circle whose radius r satisfies x² + ² = R² or r = √R² - x². The area of the cross section is А(x) = лr² = л(R² – x²). Therefore, the sphere has volume [² x (R² - x²) dx = x (R²x - 5)^² = 2(xr³ —ײ³) = ²*³ 4 - -R 3 310 / 1180
Transcribed Image Text:y A 4 DF FIGURE 3 A horizontal cross section of the pyramid is a square. y=4-x² Cross section is a semicircle of radius √4-y. FIGURE 4 -R B FIGURE 5 12-y Length √4 R 12 r= √R² ² ■ EXAMPLE 1 Volume of a Pyramid Calculate the volume V of a pyramid of height 12 m whose base is a square of side 4 m. Solution To use Eq. (1), we need a formula for the horizontal cross section A(y). Step 1. Find a formula for A(y). Figure 3 shows that the horizontal cross section at height y is a square. To find the sides of this square, apply the law of similar triangles to AABC and to the triangle of height 12-y whose base of lengths lies on the cross section: We find that s = = Step 2. Compute V 12 ² = 6" V Base 2 Height 12 12- y = V = A(y) dy = (12-y), and therefore, A(y) = s² = (12- y)². as the integral of A (y). = 6² 2(12- y) = 6s 12 12 [²²/1 (12- y)² dy = - 2/7 (12 - y) ³¹/1² This agrees with the result obtained using the formula V = Ah for the volume of a pyramid of base A and height h, since Ah = }(4²)(12) = 64. ■ ■ EXAMPLE 2 Compute the volume V of the solid in Figure 4, whose base is the region between the inverted parabola y = 4 - x² and the x-axis, and whose vertical cross sections perpendicular to the y-axis are semicircles. Solution To find a formula for the area A (y) of the cross section, observe that y = 4 - x² can be written x = ±√4-y. We see in Figure 4 that the cross section at y is a semicircle of radius r = √√4-y. This semicircle has area A(y) = r² = (4- y). Therefore, = 64 m³ A(y) dy = =-6²4- 2 π (4-y)dy = (4y - ₁²) = 4x 2 ■ In the next example, we compute volume using vertical rather than horizontal cross sections. This leads to an integral with respect to x rather than y. ■ EXAMPLE 3 Volume of a Sphere: Vertical Cross Sections Compute the volume of a sphere of radius R. Solution As we see in Figure 5, the vertical cross section of the sphere at x is a circle whose radius r satisfies x² + ² = R² or r = √R² - x². The area of the cross section is А(x) = лr² = л(R² – x²). Therefore, the sphere has volume [² x (R² - x²) dx = x (R²x - 5)^² = 2(xr³ —ײ³) = ²*³ 4 - -R 3 310 / 1180
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