MATH 116 Second Midterm Exam, Solutions

1. a) Evaluate ∫ ∫_{R} x dA as the iterated integral ∫ (∫ x dx)dy, where R is the region inside the circle x^{2} + y^{2} = 9 and over the parabola y = x^{2} + 4x + 3. Solution: If y = x^{2} + 4x + 3, then x = −2 ± \sqrt{y + 1}. Therefore,

∫_{−1}^{0} ∫_{−2−\sqrt{y+1}}^{−2+\sqrt{y+1}} x dx dy + ∫_{0}^{3} ∫_{−\sqrt{9−y^{2}}}^{−2+\sqrt{y+1}} x dx dy =

∫_{−1}^{0} \frac{1}{2} [(−2+\sqrt{y + 1})^{2}−(−2−\sqrt{y + 1})^{2}] dy+∫_{0}^{3} \frac{1}{2} [(−2+\sqrt{y + 1})^{2}−(−\sqrt{9 − y^{2}})^{2}] dy =

−4 ∫_{−1}^{0} \sqrt{y + 1} dy + \frac{1}{2} ∫_{0}^{3} (4 − 4 \sqrt{y + 1} + y + 1 − 9 + y^{2}) dy = −4 ∫_{0}^{1} \sqrt{t} dt+

\frac{1}{2} ∫_{0}^{3} (y^{2} + y − 4) dy − 2 ∫_{0}^{3} \sqrt{y + 1} dy = −4 · 2/3 t^{3/2} |_{0}^{1} + \frac{1}{2} (\frac{27}{3} + \frac{9}{2} − 4 · 3) −

2 ∫_{1}^{4} \sqrt{t} dt = −\frac{8}{3} + \frac{9}{2} + \frac{9}{4} − 6 − 2 · \frac{2}{3}(4^{3/2} − 1) = −\frac{8}{3} + \frac{27}{4} − 6 − \frac{28}{3} = −11\frac{1}{4}.

1. b) Evaluate ∫_{0}^{4} ∫_{x/2}^{2} e^{y^{2}} dydx. Solution: We change the order of integration:

∫_{0}^{2} ∫_{0}^{2y} e^{y^{2}} dx dy = ∫_{0}^{2} e^{y^{2}} 2y dy = e^{y^{2}} |_{0}^{2} = e^{4} − 1.

2. Evaluate ∫ ∫_{R} \frac{x^{2} sin[π(x^{2} + y^{2})]}{x^{2} + y^{2}} dA over the region R that lies between the circumferences x^{2} + y^{2} = 1/4, x^{2} + y^{2} = 1 and above the line y = x. Solution: In polar coordinates we get

∫_{π/4}^{5π/4} ∫_{1/2}^{1} \frac{r^{2} cos^{2}θ}{r^{2}} sin(π r^{2}) r dr dθ = ∫_{π/4}^{5π/4} \frac{1 + cos2θ}{2} dθ · ∫_{1/2}^{1} r sin(π r^{2}) dr =

(\frac{θ}{2} + \frac{sin2 θ}{4}) |_{π/4}^{5π/4} ·(\frac{−cos(π r^{2})}{2π}) |_{1/2}^{1} = [\frac{π}{2} + \frac{1}{4} (sin\frac{5π}{2} − sin\frac{π}{2} )]·\frac{1}{2π} [−cosπ + cos\frac{π}{4} ]

= [\frac{π}{2} + \frac{1}{4} · (1 − 1)] · \frac{1}{2π} [1 + \frac{\sqrt{2}}{2}] = \frac{1}{4} · \frac{2 + \sqrt{2}}{2} = \frac{2 + \sqrt{2}}{8}.

3. a) Consider the integral ∫ ∫ ∫_{D} f(x, y, z) dV, where D is the region bounded above by the hemisphere z = \sqrt{9 − x^{2} − y^{2}}, below by the xy− plane and on sides by the cylinder x^{2} + y^{2} = 1. Write the integral in cylindrical coordinates and in spherical coordinates.

Solution:

∫ ∫ ∫_{D} f(x, y, z) dV = ∫_{0}^{2π} ∫_{0}^{1} ∫_{0}^{√\overline{9−r^{2}}} f(r cosθ, r sinθ, z) r dz dr dθ =

∫_{0}^{2π} ∫_{0}^{arcsin1/3} ∫_{0}^{3} f(ρ cosθ sinφ, ρ sinθ sinφ, ρ cosφ) ρ^{2} sinφ dρ dφ dθ+

∫_{0}^{2π} ∫_{arcsin1/3}^{π/2} ∫_{0}^{1/sinφ} f(ρ cosθ sinφ, ρ sinθ sinφ, ρ cosφ) ρ^{2} sinφ dρ dφ dθ.

For spherical coordinates we use the following calculations:

ρ = 3, r = 1 ⇒ sinφ_{0} = 1/3. Also, x^{2} + y^{2} = 1 ⇒ ρ sinφ = 1.

3. b) Find the volume of the region enclosed by the surfaces z = x^{2} + y^{2} and z = (x^{2} + y^{2} + 1)/2.

Solution: In cylindrical coordinates the surfaces z = r^{2} and z = (r^{2} + 1)/2 intersect when r = 1. Therefore,

V = ∫_{0}^{2π} ∫_{0}^{1} ∫_{r^{2}}^{r^{2}+1} r dz dr dθ = 2π ∫_{0}^{1} r (\frac{r^{2} + 1}{2} − r^{2}) dr = π ∫_{0}^{1} r(1 − r^{2}) dr =

π(1/2 − 1/4) = π/4.

4. Let D be the region in the first quadrant bounded by the hyperbolas

xy = 1, xy = 9 and the lines y = x, y = 4x. Evaluate ∫ ∫_{D} (\sqrt{y/x} + \sqrt{xy}) dA.

Solution: We use the substitution u = xy and v = y/x. Then in new coordinates u, w the region D is the rectangle 1 ≤ u ≤ 9, 1 ≤ v ≤ 4. Here,

\frac{∂(u, v)}{∂(x, y)} = | \begin{matrix}y x \\ −\frac{y}{x^{2}} \frac{1}{x} \end{matrix} | = y/x + y/x = 2v. Therefore, \frac{∂(x, y)}{∂(u, v)} = (2v)^{−1} and

∫ ∫_{D} (\sqrt{y/x}+\sqrt{xy}) dA = ∫_{1}^{9} ∫_{1}^{4} \frac{\sqrt{v} + \sqrt{u}}{2v} dv du = ∫_{1}^{9} (\sqrt{v} + \frac{1}{2}\sqrt{u} lnv) |_{v=1}^{v=4} du =

∫_{1}^{9} (1 + \frac{ln4 − ln1}{2} \sqrt{u}) du = (u + ln2 · \frac{2}{3} u^{3/2}) |_{1}^{9} = 8 + ln2 · \frac{2}{3} · 26 = 8 + \frac{52}{3} ln2.

5. The region R is bounded by the planes 4x + y − 2z = 0, x − y + 2z = 0, y = 0, and x − y − 3z + 5 = 0. Use the transformation x = u − 2w, y = 2v, z = 2u + v + w to evaluate ∫ ∫ ∫_{R} y dV.

Solution: Here, 0 = 4x + y − 2z = 4u − 8w + 2v − 4u − 2v − 2w = −10w, 0 = x − y + 2z = u − 2w − 2v + 4u + 2v + 2w = 5u, 0 = y = 2v, and 0 = x−y −3z +5 = u−2w −2v −6u−3v −3w +5 = 5(1−u−v −w). Therefore the preimage of R under the given transformation is the simplex G defined by the conditions u ≥ 0, v ≥ 0, w ≥ 0, u + v + w ≤ 1. The Jacobian of transformation is

\frac{∂(x, y, z)}{∂(u, v, w)} = | \begin{matrix}1 0 − 2 \\ 0 2 0 \\ 2 1 1 \end{matrix} | = 2 | \begin{matrix}1 − 2 \\ 2 1 \end{matrix} | = 10.

Thus,

∫ ∫ ∫_{R} y dV = ∫_{0}^{1} ∫_{0}^{1−u} ∫_{0}^{1−u−v} 2v·10 dw dv du = 20 ∫_{0}^{1} ∫_{0}^{1−u} v(1−u−v)dv du =

20 ∫_{0}^{1}[(1 − u) v^{2}/2 − v^{3}/3] |_{v=0}^{v=1−u} du = 20 ∫_{0}^{1}(1 − u)^{3} (1/2 − 1/3) du = \frac{20}{6} ∫_{0}^{1} t^{3} dt =

10/3 · 1/4 = 5/6.