Figure 1: The region R in question I.

Mathematics 136, section 2 – Advanced Placement Calculus Solutions for Exam 3 – December 2, 2009

I. Let R be the region in the plane bounded by y = e^{x}, y = x/2, x = 0, and x = 1.

A) (5) Sketch the region R.

Solution. See Figure 1. B) (5) Find the area of R.

Solution: We have

A = ∫_{0}^{1} e^{x} − x/2 dx = e^{x} − x^{2}/4|_{0}^{1} = e − 1/4 − 1 = e − 5/4 ≐ 1.47.

C) (10) Find the volume of the solid generated by rotating R about the x-axis. Solution: The cross-sections are washers with outer radius e^{x} and inner radius x/2, so

V = ∫_{0}^{1} π(e^{x})^{2} − (x/2)^{2} dx
= π ∫_{0}^{1} e^{2x} − x^{2}/4 dx
= π (e^{2x}/2 − x^{3}/12|_{0}^{1})
= π(e^{2}/2 − 1/12 − 1/2)
= π(e^{2}/2 − 7/12).

D) (5) Set up, but do not evaluate, an integral or integrals to compute the volume of the solid generated by rotating R about the y-axis.

Solution: The cross-sections are different for different ranges of y: disks with radius x = 2y for 0 ≤ y ≤ 1/2, disks with radius 1 for 1/2 ≤ y ≤ 1, and washers with outer radius 1 and inner radius x = ln(y) for 1 ≤ y ≤ e. So the volume would be computed as the sum:

V = ∫_{0}^{1/2} π(2y)^{2} dy + ∫_{1/2}^{1} π(1)^{2} dy + ∫_{1}^{e} π((1)^{2} − (ln(y))^{2}) dy.

II. A) (10) Integrate by parts: ∫ xe^{−3x} dx.

Solution: Let u = x and dv = e^{−3x} dx. Then du = dx and v = \frac{−1}{3}e^{−3x}. Applying the parts formula,

∫ xe^{−3x} dx = \frac{−xe^{−3x}}{3} + \frac{1}{3} ∫ e^{−3x} dx = \frac{−xe^{−3x}}{3} − \frac{e^{−3x}}{9} + C.

B) (10) Does the integral ∫_{0}^{∞} xe^{−3x} dx converge? If so, find its value.

Solution: The integral does converge to the value \frac{1}{9}:

∫_{0}^{∞} xe^{−3x} dx = lim_{b→∞} ∫_{0}^{b} xe^{−3x} dx

= lim_{b→∞} ( \frac{−xe^{−3x}}{3} − \frac{e^{−3x}}{9} |_{0}^{b})

= lim_{b→∞} (\frac{−be^{−3b}}{3} − \frac{e^{−3b}}{9} + 0 + \frac{1}{9}) .

Now,

lim_{b→∞} \frac{e^{−3b}}{9} = lim_{b→∞} \frac{1}{9e^{3b}} = 0.

Moreover, by L’Hopital’s Rule,

lim_{b→∞} \frac{−be^{−3b}}{3} = lim_{b→∞} \frac{−b}{3e^{3b}}  (an ∞/∞ form)

= lim_{b→∞} \frac{−1}{9e^{3b}}

= 0.

Hence the integral converges to \frac{1}{9}.

III. (15) A water tank has the shape of an inverted cone (that is, the “point” of the cone is at the bottom) with height 10 meters and radius 3 meters. If the tank is full of water (density 1000kg per cubic meter), find the work done in pumping all the water out the top of the tank. The acceleration of gravity is 9.8 meters per second squared.

Solution: Consider the slice of the water, thickness Δy, originally at height y from the bottom – approximately a cylinder with radius 3y/10. The work to lift the slice is

W_{slice} = π(3y/10)^{2}Δy · 1000 · 9.8 · (10 − y).

Figure 2: The slope field and solution in V.

Summing over the slices and taking a limit:

W = 9800π ∫_{0}^{10} \frac{9y^{2}}{10} − \frac{9y^{3}}{100} dy = 9800π ( \frac{3y^{3}}{10} − \frac{9y^{4}}{400}|_{0}^{10}) = 735000π.

IV. (15) Set up and evaluate the integral to compute the arclength of the parabola y = 4x^{2}, 0 ≤ x ≤ 1. (Use the table to evaluate.) Solution: y′ = 8x, so

L = ∫_{0}^{1} \sqrt{1 + (8x)^{2}} dx.

To evaluate, we use the substitution u = 8x and # 21 in the table of integrals:

= \frac{1}{8} ( \frac{8x}{2}\sqrt{1 + (8x)^{2}} + \frac{1}{2} ln(8x + \sqrt{1 + (8x)^{2}}|_{0}^{1}) = \frac{\sqrt{65}}{2} + \frac{1}{16} ln(8 + \sqrt{65}).

V. All parts of this question refer to the differential equation

\frac{dy}{dx} = (y − 1)(y − 3)/4

A) (10) Sketch the slope field of this equation, showing the slopes at points on the lines y = 0, 1, 2, 3, 4, 5, and −3 ≤ x ≤ 3. Solution: See Figure 2.

B) (5) On your slope field, sketch the graph of the solution of the equation with y(0) = 2.

Solution: See Figure 2. C) (10) This is a separable equation; find an explicit formula for the solution satisfying the

initial condition y(0) = 2.

Solution: We separate variables and integrate using partial fractions for the y-integral:

∫ \frac{dy}{(y − 1)(y − 3)} = ∫ \frac{1}{4} dx
∫ \frac{−1/2}{y − 1} + \frac{1/2}{y − 3} = \frac{x}{4} + c
−\frac{1}{2} ln |y − 1| + \frac{1}{2} ln |y − 3| = \frac{x}{4} + c
ln |\frac{y − 3}{y − 1}| = \frac{x}{2} + c
\frac{y − 3}{y − 1} = ke^{−x/2} where k = ±e^{c}
y − 3 = k(y − 1)e^{−x/2}
y = \frac{−ke^{x/2} + 3}{−ke^{x/2} + 1}.

To get y(0) = 2, we must have 2 = \frac{3−k}{1−k} so 2 − 2k = 3 − k, or k = −1. The solution can be written as

y = \frac{e^{x/2} + 3}{e^{x/2} + 1}.

Extra Credit (10) Set up the integrals to find the coordinates of the centroid of the region in the first quadrant inside the ellipse \frac{x^{2}}{4} + y^{2} = 1. You do not need to evaluate.

The region is bounded by y = \sqrt{1 − x^{2}/4}, x = 0, x = 2, and the x-axis. The coordinates of the centroid would be computed by

\overline{x} = \frac{∫_{0}^{2} x\sqrt{1 − x^{2}/4} dx}{∫_{0}^{2} \sqrt{1 − x^{2}/4} dx}

and

\overline{y} = \frac{∫_{0}^{2} \frac{1}{2}(\sqrt{1 − x^{2}/4})^{2} dx}{∫_{0}^{2} \sqrt{1 − x^{2}/4} dx}.

(These are approximately (\overline{x}, \overline{y}) ≐ (.85, .42). See Figure 3.)

Figure 3: The region and the approximate location of the centroid. 5