Solutions and Commentary on MATH 251 (22,23,24 ) [Fall 2020], Dr. Z. , Exam 2, Monday, Nov. 23, 2020, 8:40-10:40am

Types: Number, Function of variable(s), 2D vector of numbers, 3D vector of numbers, 2D vector of functions, 3D vector of functions, equation of a plane, parametric equation of a line, equation of a line, equation of a surface, equation of a line, DNE (does not exist), parametric equation of surface, double integral of an abstract function.

1. (10 pts.) Compute the line integral

∫_{C} yz dx + (xz + z) dy + (xy + y + 1) dz ,

over the path

r(t) = ⟨e^{t^{3}}, t^{2}e^{t^{4}}, te^{t^{7}}⟩ , 0 ≤ t ≤ 1 .

Explain!

The type of the answers is: Number.

ans. e + e^{2} + e^{3}

Comment: This was meant to be done the clever way. People who did it directly, and then used Maple to evaluate it, and got the right answer, still got full credit, but in the Final I may specifiy ”do not use Maple” (or any other software). Also I gave full credit for the answer in decimals. The perfect answer should be in terms of e. In the final exam I will insist on it (but will mention it explicitly).

Sol. Whenever the vector field is conservative, then it is much easier to use the Fundamental Theorem of Line Integrals. Taking the curl(F) gives < 0, 0, 0 > (you do it!) hence there is a potential function f(x, y, z) such that grad(f) =< yz, xz+z, xy+y+1 >. i.e

f_{x} = yz , f_{y} = xz + z , f_{z} = xy + y + 1 .

From f_{x} = yz we get f(x, y, z) = xyz + ϕ(y, z) where ϕ(y, z) is to be determined. Using f_{z} = xz + z, we get xz + ϕ_{y}(y, z) = xz + z, hence ϕ_{y}(y, z) = z. Intergating with resepct to y, we get ϕ(y, z) = yz + ψ(z) where ψ(z) is to be determined. We currently have f(x, y, z) = xyz + yz + ψ(z). Using f_{z} = xy + y + 1, we get xy + y + ψ′(z) = xy + y + 1 so ψ′(z) = 1 and so ψ(z) = z + C (but C is not important for this problem) so

Crucial First Step: The potential function if f(x, y, z) = xyz + yz + z.

Comment: At this step it is a good idea to check That grad(f) is indeed < yz, xz + z, xy + y + 1 >. Do it!

Now we almost done. By the Fundamental theorem of Line Integrals the value is f(End) − f(Start).

The starting point is r(0) = (1, 0, 0), the ending point is r(1) = (e, e, e), so the final answer is

f(e, e, e) − f(1, 0, 0) = e^{3} + e^{2} + e − 0 = e^{3} + e^{2} + e .

2. (10 points) By changing the order of integration, if necessary, evaluate the doubleintegral

∫_{0}^{5} ∫_{(y/5)^{1/3}}^{1} sin x^{4} dx dy

The type of the answer is: Number

ans. \frac{5}{4}(1 − cos 1)

Sol. The region of integration is, in Type II format

D = {(x, y) : 0 < y < 5, (y/5)^{1/3} < x < 1} .

Plotting it, and realizing that the curve x = (y/5)^{1/3} can be written in the usual format as y = 5x^{3} (solve for y in terms of x), the same region in Type I format is

D = {(x, y) : 0 < x < 1, 0 < y < 5x^{3}} .

This leads to the iterated integral

∫_{0}^{1} ∫_{0}^{5x^{3}} sin x^{4} dy dx .

The inner integral is

∫_{0}^{5x^{3}} sin x^{4} dy = sin x^{4} ∫_{0}^{5x^{3}} dy =
= sin x^{4}(y|_{0}^{5x^{3}}) = 5x^{3} sin x^{4} .

The outer integral is

∫_{0}^{1} 5x^{3} sin x^{4} dx .

Doing the substituion u = x^{4} we get that this

\frac{5}{4} ∫_{0}^{1} sin u du = \frac{5}{4}(− cos u)|_{0}^{1}= \frac{5}{4}(− cos 1 − (− cos 0)) = \frac{5}{4}(1 − cos 1) .

Comment: Once again, I forgot to state ”do not use Maple”. People who did the clever change of order, and then used Maple got full credit. In the Final I may add the restrictions, ”do not use a computer” and ”explain everything.

3. (10 points) Find the equation of the tangent plane at the point (1, 1, 1) to the surface given parametrically by

x(u, v) = u^{3}v , y(u, x) = uv , z(u, v) = uv^{3} , −∞ < u < ∞ , −∞ < v < ∞ .

Express you answer in explicit form, i.e in the format z = ax + by + c.

The type of the answer is: Equation of a Plane

ans. z = −x + 4y + 2

Sol. The parametric equation of the surface in vector notation is

r(u, v) = ⟨u^{3}v , uv , uv^{3}⟩ .

We have

r_{u} = ⟨3u^{2}v , v , v^{3}⟩ .
r_{v} = ⟨u^{3} , u , 3uv^{2}⟩ .

Now it is time to plug it in. But first we need to find the values of u and v that correspond to (1, 1, 1). This can be easily seen to be u = 1, v = 1.

r_{u}(1, 1) = ⟨3 , 1 , 1⟩ .
r_{v}(1, 1) = ⟨1 , 1 , 3⟩ .

In order to find a Normal vector, we find the cross product r_{u} × r_{v}

N = det ( \begin{matrix}i & j & k \\ 3 & 1 & 1 \\ 1 & 1 & 3 \end{matrix} )

= i(1 · 3 − 1 · 1) − j(3 · 3 − 1 · 1) + k(1 · 3 − 1 · 1) = 2i − 8j + 2k = ⟨2, −8, 2⟩ .

The point, (x_{0}, y_{0}, z_{0}) is (1, 1, 1), so using

N.⟨x − x_{0}, y − y_{0}, z − z_{0}⟩ = 0 ,

we get

⟨2, −8, 2⟩.⟨x − 1, y − 1, z − 1⟩ = 0 .

Dividing by 2 this gives

(x − 1) − 4(y − 1) + (z − 1) = 0 .

Simplifying, and solving for z gives z = −x + 4y + 2.

4. (10 points) Let f(x, y, z) = e^{cos x^{2} + sin xyz+cos xz}, and let

F = ⟨\frac{∂f}{∂x} , \frac{∂f}{∂y} , \frac{∂f}{∂z} ⟩

Let C be the curve

r(t) = ⟨cos t, t, sin t⟩ , 0 ≤ t ≤ 2π .

Find the value of the line-integral

∫_{C} F.dr .

Explain! Just giving the answer will give you no credit.

The type of the answer is: Number

ans. 0

Comment: The original version claimed that C is closed. I wrote email to everyone telling them to delete the word ”closed”. It so happens that the answer is the same either way, 0, and I gave full credit to both versions.

Sol. This obviously called for the Fundanental Theorem of Line Inegrals. F is grad(f) so the answer is f(End) − f(Start). The staring point is ⟨1, 0, 0⟩, and the end point is ⟨1, 2π, 0⟩. So the answer is

e^{cos 1 +1} − e^{cos 1 +1} = 0.

5. (10 points) Evaluate the triple integral

∫_{R} (x^{2} + y^{2} + z^{2})^{3} dx dy dz ,

where R is the region in 3D space given by

{(x, y, z) | x^{2} + y^{2} + z^{2} ≤ 1 , x, y, z ≥ 0} .

The type of the answer is: Number

ans. \frac{π}{18} Sol. This calls for spherical coordinates. Recall that dx dy dz = ρ^{2} sin ϕ dρdϕdθ. So the integrand is (recall that x^{2} + y^{2} + z^{2} = ρ^{2}) (ρ^{2})^{3}ρ^{2} sin ϕ. We also need the region in these coordinates. It is

\{(ρ, ϕ, θ) : ρ < 1 , 0 < ϕ < \frac{π}{2} , 0 < θ < \frac{pi}{2}\} .

Hence the requred integral is

∫_{0}^{π/2} ∫_{0}^{π/2} ∫_{0}^{1} ρ^{8} sin ϕ dρdϕdθ .

This equals

(∫_{0}^{1} ρ^{8}dρ) (∫_{0}^{π/2} sin ϕdϕ) (∫_{0}^{π/2} dθ) = \frac{1}{9} · 1 · \frac{π}{2} = \frac{π}{18} .

VERY FREQUENT ERROR: Many people did not see the x, y, z > 0 and integrated over the whole sphere, with limits of integration in ϕ from 0 to π and in θ from 0 to 2π. READ THE QUESTION!. I gave lots of partial credit for people who made this error, but I am disappointed that so many people did not notice it. I gave you half an hour to check and double-check! I am willing to bet that many people used the whole two hours to do problems, and checked very quickly, and not at all.

Warning: A similar error (not reading the question properly and doing a different problem), in the final exam will give you zero points.

Comment: If you are really adbanced, you could have said that since the integrand is symmetric with respect to exchanging x and −x, the desired value is one-eighth of the answer of integrating over the whole spehere, and if you did over tge whole sphere, (as many people did) you get \frac{4π}{9} , and to get the desired answer you divide by 8. But this argument is ‘dangenrous’ and is only applicable for symmetric integrands.

6. (10 points) Evaluate the double integral

∫_{−3}^{0} ∫_{0}^{√\overline{9−x^{2}}}(x^{2} + y^{2})^{2} dy dx

The type of the answer is: Number

ans. \frac{243}{4}

Sol. The region, in type I notation (using the usual, rectangular, coordinates) is

\{(x, y) : −3 < x < 0, 0 < y < \sqrt{9 − x^{2}}\}

A little diagram shows that it is the quarter circle center origin, radius 3 that lies to the left of y-axis and above the x-axis, in other words, the quarter circle of radius 3 that lies in the second quadrant.

A little diagram shows that the region, expressed in the polar language is

\{(r, θ) : 0 < r < 3, \frac{π}{2} < θ < π\} .

Converting to polar cooridnates we have that the integrand is r^{4} · r = r^{5}.

So we have to compute

∫_{0}^{3} ∫_{π/2}^{π} r^{5} dθ dr = (∫_{0}^{3} r^{5} dr) · (∫_{π/2}^{π} dθ) = (\frac{r^{6}}{6} |_{0}^{3}) · (π − \frac{π}{2} ) =

\frac{3^{6}π}{6 · 2} = \frac{243}{4}π .

Comment: Maple is smart enough to do it directly. Since I allowed Maple, and did not speficy not to use it, I gave full credit to people who got the right answer using Maple. If such a problem will come up in the Final Exam, I would probably say ”...by converting to polar coordinates and doing it completely by hand...“

7. (10 points altogether) Decide whether the following limits exist. If it does find them. If it does not Explain why not?

(a) (2 points) lim_{(x,y)→(π/2,π/2)} \frac{cos x + sin x}{x + y} , (b) (2 points) lim_{(x,y)→(0,0)} \frac{x^{2} − y^{2}}{x − y} ,

(c) (2 points) lim_{(x,y)→(0,0)} \frac{x − y}{x^{2} − y^{2}} , (d) (4 points) lim_{(x,y)→(1,1)} \frac{x + y − 2}{2x + y − 3} ,

ans. (a) \frac{1}{π} (b) 0 (c) DNE (d) DNE

Sol. (a) Just plug-it-in. There are no issues and the answer is \frac{1}{π}

(b): Now we get 0/0, but simplifying we get x + y. Now there is no issue in plugging-it-in and we get 0.

(c) Now we simplify and get 1/(x + y). Now we plug-it-in, and get 1/0, so obviously DNE.

(d) This is the hardest, and it is similar to a problem in a previous exam. I am still very disappointed that quite a few peple did not get it. Once again plugging-it-in gives 0/0. SImplifying fails, so we explore the function from various direction, and see the limits. If you use the line (x, y) = (1 + t, 1) parallel to the x axis, and let t go to 0 you would get \frac{1}{2} so the limit in the sense of calc1 is \frac{1}{2}. If you use the line (x, y) = (1, 1 + t), and let t go to 0 you would get 1. Since there is no agreement, the limit (in the sense of calc3) does not exist).

Comment: A more general argument would be to consider the line through (1, 1) with slope m y = 1 + m(x − 1) and get that the limit as m goes to 0 is \frac{m+1}{m+2}. Since it depends on the slope there is no agreeement.

8. (10 points) Compute the line integral ∫_{C} f ds where

f(x, y, z) = xyz

and C is the line segment starting at (0, 0, 0) and ending at (1, 2, −3)

The type of the answer(s) is: Number ans. −\frac{3}{2} \sqrt{14} .

Sol.: We first need a parametric representation r(t). Recall that s line segment from P to Q is P + t(Q − P), where t goes from 0 to 1. In this case it is

r(t) = ⟨t, 2t, −3t⟩ .

We have

r′(t) = ⟨1, 2, −3⟩ .

So

ds = |r′(t)| dt = \sqrt{1^{2} + 2^{2} + (−3)^{2}} dt .

Replacing x, y and z in f(x, y, z) by t, 2t, −3t, respectively, gives (t)(2t)(−3t) = −6 t^{3}, so the desired integral is

∫_{0}^{1} −6t^{3}\sqrt{14} dt = −\sqrt{14}\frac{6t^{4}}{4} |_{0}^{1} = −\sqrt{14}\frac{3}{2} .

9. (10 points) Compute the vector-field surface integral ∫ ∫_{S} F.dS if F is

F = ⟨z, z, x⟩ ,

and S is the oriented surface

z = 9 − x^{2} − y^{2} , x ≥ 0, y ≥ 0, z ≥ 0

with downward pointing normal.

The type of the answer is Number

ans. −\frac{693}{5} .

Sol.: In this class we are only supposed to know how to compute vector-field surface integrals over surfaces that are given in explicit format z = g(x, y) for some function of x and y. The more general case of surfaces defined parametrically is complicated, and can be found in the book, and this problem can be done that way using the parametrization ⟨x, y, 9 − x^{2} − y^{2}⟩ and thinking of x and y as parameters. But not recommended.

The good way of doing it is using the formula given in p.4 of handout 16.5

∫ ∫_{Ω} (−P \frac{∂g}{∂x} − Q\frac{∂g}{∂y} + R) dA ,

where Ω is the “floor-plan”. Sometimes it is given directly (like in the last problem of handout 16.5), but sometimes, like in this problem, you have to figure it out yourself.

The way to find Ω is to set g(x, y) = 0, so in this problem it is the disc x^{2} +y^{2} ≤ 9, i.e. the circle of radius 3 and center the origin. But note the x, y > 0, so the relevant ‘floor-plan’ is the quarter disc in the first quadrant. Here we have

P = z , Q = z , R = x ,

\frac{∂g}{∂x} = −2x , \frac{∂g}{∂y} = −2y .

So what we need (with the usual upward direction) is

∫ ∫_{Ω} (−(z)(−2x) − z(−2 ∗ y) + x)dA = ∫ ∫_{Ω} (2x + 2y)z + x)dA ,

But don’t forget to replace z by 9 − x^{2} − y^{2}, getting

∫ ∫_{Ω} (2x + 2y)(9 − x^{2} − y^{2}) + x)dA .

and in usual (rectangular) coordinates it is

∫_{0}^{3} ∫_{0}^{√\overline{9−x^{2}}} (2x + 2y)(9 − x^{2} − y^{2}) + x)dydx .

One way is to use Maple,

int(int((2*x+2*y)*(9-x**2-y**2)+x,y=0..sqrt(9-x**2)),x=0..3);

and immediately get \frac{693}{5}, but converting to polar coordinates gives

∫_{0}^{3} ∫_{0}^{π/2}(2r(cos θ + sin θ)(9 − r^{2}) + r cos θ)r dθ

and once again you are welcome to use Maple (if you wish).

Finally since it says downward direction, we have to multiply it by −1, getting the answer.

Comments: 1. Quite a few people got the integrand correctly, but possibly ‘adpating’ the problem in handout 16.5, took the region to be 0 < x, y < 3 or other ones. Be careful!

2. Many people did not notice the downward word, and did not multiply by −1. This time I was lenient, and only took off two points, but in the Final, you would get zero points. Read the question! I gave you half an hour to check!

10. (10 points) Find the point on the plane x + 2y + 3z = 18 where the function f(x, y, z) = xyz is as large as possible.

The type of the answer is: POINT

ans. (6, 3, 2)

Sol. We use Lagrange multipliers. Here the goal function is f(x, y, z) = xyz and the constraint is x + 2y + 3z = 18 so g(x, y, z) = x + 2y + 3z.

We need

grad(f) = λgrad(g)

So

< langleyz, xz, xy⟩ = λ⟨1, 2, 3⟩ .

Spelling this out, and adding the constraint we have the system

{yz = λ, xz = 2λ, xy = 3λ, x + 2y + 3z = 18}

Dividing the yz = λ by xz = 2λ we get \frac{y}{x} = \frac{1}{2}. Dividing the yz = λ by xy = 3λ we get \frac{z}{x} = \frac{1}{3}. So y = \frac{1}{2}x, z = \frac{1}{3}x. Using x + 2y + 3z = 18 we get

x + 2(\frac{1}{2}x) + 3(\frac{1}{3}x) = 18 ,

So

3x = 18 ,

so x = 6. Going back to y = \frac{1}{2}x, and z = \frac{1}{3}x we get y = 3 and z = 2.

The solution to the system is x = 6, y = 3, z = 2 (and we really don’t care about λ that happens to be 6).

So the desired answer is that the point is (6, 3, 2).

Comment: I was very disappointed that many people, including some of the best, gave the anser 36. This is the value, but I asked for the location (point).

READ THE QUESTION!

This time I was lenient and gave lots of partial credit for people who did it correctly except that the put 36 as the answer (and of course had to find the x, y and z in the process, but I gave you half an hour to check that you answered the right question not a different one. In the Final exam you would get zero points!