Pre-Calc Homework Solutions 301

Pre-Calc Homework Solutions 301 - Section 7.3 2. In each...

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Unformatted text preview: Section 7.3 2. In each case, the width of the cross section is w (a) A (b) A (c) A (d) A A(x) r 2, where r s 2, where s s 2, where s w , so A(x) 2 w 2 2 301 2 x. x. 8. A cross section has width w w 2 2 sec x tan x. (a) A(x) 4x. /3 r2 (sec x w, so A(x) w 2 w 2 4 (sec x tan x)2, and , so A(x) w 2 2 2x. V /3 4 /3 tan x)2 dx 2 sec x tan x tan x 1 x 2 /3 /3 4 3 2 w (see Quick Review Exercise 5), so 3 4 4 (sec2 x tan2 x) dx /3 /3 (2 x)2 3x. 4 tan x tan x 3 3 2 sec x sec x 2 2 x /3 3. A cross section has width w A(x) 4 2 x and area 2 s2 x2 w 4 0 2 2x. The volume is 16. (2 x 2) (1 2x 2 3 x 3 2 2 2 2x dx 0 6 3 2 6 4. A cross section has width w and area A(x) 1 x2 2 2 2 2x 2 (b) A(x) V 6 . (sec x tan x)2, and r 2 1 w 2 2 s2 /3 /3 w2 (sec x x ) . The volume is 1) dx 1 tan x)2 dx, which by same method as 2 . 3 (1 1 x ) dx 2 2 (x 1 4 in part (a) equals 4 3 9. A cross section has width w r2 w 2 2 4 1 5 x 5 16 . 15 x 1 5y 2 and area 5. A cross section has width w A(x) 1 2 1 x 2 and area 1 1 2 0 5 4 y dy 4 y5 5 4 y . The volume is 4 2 8 . 0 s2 w2 x 2) dx 4(1 1 x 2). The volume is (1 1 4(1 1 4 x 2) dx 2 1 4 x 1 3 x 3 16 . 3 10. A cross section has width w 1 2 s 21 1 2 w 2 2 1 1 1 y 2 and area 2(1 y 2). The volume is 2y 1 3 y 3 8 . 3 6. A cross section has width w A(x) 1 x 2 and area 1 1 2(1 1 y 2) dy s2 2 w 2 2 2(1 1 x 2). The volume is x ) dx 2 2(1 1 x ) dx 2 (1 1 2 x 1 3 x 3 8 . 3 11. (a) The volume is the same as if the square had moved without twisting: V Ah s 2h. (b) Still s 2h: the lateral distribution of the square cross sections doesn't affect the volume. That's Cavalieri's Volume Theorem. 12. Since the diameter of the circular base of the solid extends from y 12 2 7. A cross section has width w 3 2 w 2 sin x. (a) A(x) V 0 4 3 sin x, and 6 to y 12, for a diameter of 6 and a radius 3 sin x dx 3 0 of 3, the solid has the same cross sections as the right circular cone. The volumes are equal by Cavalieri's Theorem. sin x dx cos x 0 3 2 (b) A(x) V 0 3. s2 w2 4 sin x, and 4 0 13. The solid is a right circular cone of radius 1 and height 2. V 4 cos x 0 1 Bh 3 1 ( r 2)h 3 1 ( 12)2 3 2 3 4 sin x dx sin x dx 8. 14. The solid is a right circular cone of radius 3 and height 2. V 1 Bh 3 1 ( r 2)h 3 1 ( 32)2 3 6 ...
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