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# Solutions to Areas and Volumes Practice Problems for AP Calculus

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By McGraw-Hill Professional
Updated on Oct 24, 2011

Practice problems for these solutions can be found at: Areas and Volumes Practice Problems for AP Calculus

Part A—The use of a calculator is not allowed.

1.
1. Since F is decreasing on the interval [3, 5].
2. At t =3, F has a maximum value.
3. F '(x )= f (x ), F '(x ) is increasing on (4, 5) which implies F ≤ (x ) > 0. Thus F is concave upwards on (4, 5).
2. See Figure 12.9-1.
3. See Figure 12.9-2.
4. Set 2x – 6=0; x =3 and

5. See Figure 12.9-3.
6. See Figure 12.9-4.
7. See Figure 12.9-5.
8. See Figure 12.9-6.
9. Intersection points: 4= y2 y = ± 2.

You can use the symmetry of the region and obtain the area An alternative method is to find the area by setting up an integral with respect to the x-axis and expressing x = y2 as .

10. See Figure 12.9-7.
11. See Figure 12.9-8.
12. Using the Disc Method:

13. See Figure 12.9-9.
14. Using the Disc Method:

Note: You can use the symmetry of the region and find the volume by

15. Volume of solid by revolving R:
16. You can verify your result by evaluating

The result is 96π.

Part B—Calculators are allowed.

17. See Figure 12.9-10.
18. Using the Washer Method: Points of intersection: Set x3 =x2 x3x2 =0 x2(x – 1)=0 or x =1.

Step 2.

Step 3. Enter and obtain .

19. See Figure 12.9-11.
20. Step 1.

Let s = a side of an equilateral triangle .

Step 2. Area of a cross section:

Step 3.

Step 4. Enter

21. See Figure 12.9-12.
22. Step 1. Using the Washer Method: Points of Intersection:

Step 2.

Step 3.

23. See Figure 12.9-13.
24. Step 1: Using the Washer Method:

y =8, y =x3

Step 2:

25. See Figure 12.9-14.
26. Using the Washer Method:

27. See Figure 12.9-15.
28. Step 1: Using the Disc Method: Radius =(8 – x3).

Step 2:

29. See Figure 12.9-16.
30. Using the Disc Method:

31. Area of RectI= f (2)Δx1=(2.24)(4)=8.96.

Area of RectII= f (6)Δx2=(3.61)(4)=14.44.

Area of RectIII= f (10)Δx3=(4.58)(4)=18.32.

Total Area =8.96+14.44+18.32=41.72.

The area under the curve is approximately 41.72.

32. The area enclosed by the curve is the upper half of an ellipse.

The negative simply indicates that the area has been swept from right to left, rather than left to right, and so may be ignored. The area enclosed by the curve is 3π.

33. Differentiate to find
34. Find and Square each derivative.

35. Square r2 =4+8 sin θ +4 sin2θ. The area
36. The acceleration vector for an object moving in the plane is et, et . Find the position of the object at t =1 if the initial velocity is v0 = 3, 1 and the initial position of the object is at the origin.
37. The acceleration of the object is known to be a = et, et = – eti +etj. Integrate to find the velocity. v = –eti +etj +C and since the initial velocity is v0 = 3, 1, v0 = – i + j +C =3i + j and C =4i. The velocity vector is veti +etj + 4i =4 – et i +etj. Integrate again to find the position vector

s =(4te t )i +e tj +C. The initial position at the origin means that s0 =(4.0 – e0)i +e0j +C = – i + j +C =0 and therefore, C =ij. The position vector s =(4tet +1)i + (et – 1) j can be evaluated at t =1 to find the position as (5 – e)i +(e – 1) j =5 – e, e – 1.

38. (See Figure 12.10-1.)
39.

40. 27. See Figure 12.10-2.

Differentiate both sides:

41. See Figure 12.10-3.
42. Step 1. Distance Formula

Step 2.

Step 3. Use the [Zero] function and obtain x =2 for y2.

Step 4. Use the First Derivative Test. (See Figures 12.10-4 and 12.10-5.) At x =2, L has a relative minimum. Since at x =2, L has the only relative extremum, it is an absolute minimum.

Step 5.

43.
1. v(2)=2 cos(22 +1)=2 cos(5)= 0.567324
2. Since v(2) > 0, the particle is moving to the right at t =2.

3. a(t)=v '(t)
4. Enter d(x * cos(x^2+1), x )|x =2 and obtain 7.95506.

Thus, the velocity of the particle is increasing at t =2, since a(2) > 0.

44. See Figure 12.10-6
1. Point of Intersection: Use the [Intersection] function of the calculator and obtain (0.517757, 0.868931).
2. 0, 0.51775) and obtain 0.304261. The area of the region is approximately 0.304.

3. Step 1. Using the Washer Method:

Step 2.

The volume of the solid is approximately 1.167.

45. Convert to a parametric representation with x = r cos θ =5 cos θ cos 2 θ and y =r sin θ =5 cos 2θ sin θ. Differentiate with respect to θ.
46. The slope of the tangent line is zero, including a horizontal tangent.

47. can be integrated with a partial fraction decomposition. Since Therefore,

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