Final Homework Answers

Problem 1.  (a) First, you likely need to draw a picture of the two bounding surfaces (the part of the sphere and the cone).  From this picture one can see that the sphere is “on top” and the cone is “on the bottom.”  This lets us describe the solid region using the variable z first.

The solid region can be described by:

\sqrt{x^2+y^2} \leq z \leq \sqrt{4-x^2-y^2}

\displaystyle -\sqrt{2-x^2} \leq y \leq \sqrt{2-x^2}

\displaystyle -\sqrt{2} \leq x \leq \sqrt{2}

The x and y descriptions / inequalities are obtained from our picture and from solving the equations \sqrt{x^2+y^2} = \sqrt{4-x^2-y^2} to find that x^2+y^2 = 2.

The volume is then given by

\displaystyle \text{Vol}(S) = \iiint_S \, dV = \int_{-\sqrt{2}}^{\sqrt{2}} \, \int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}\,\int_{\sqrt{x^2+y^2}}^{\sqrt{4-x^2-y^2}} \, dz\,dy\,dx.

(b) To actually evaluate this volume, one should change variables.  Switching either to cylindrical or to spherical coordinates can make this problem do-able.  Here is the cylindrical coordinate way.

(Cylindrical Coordinates).  The region S can be described by the inequalities

\displaystyle r \leq z \leq \sqrt{4-r^2}

\displaystyle 0 \leq r \leq \sqrt{2}

\displaystyle 0 \leq \theta \leq 2\pi.

We then apply our change-of-variables formula to compute

\displaystyle \text{Vol}(S) = \iiint_S \, dV = \int_0^{2\pi} \, \int_0^{\sqrt{2}} \, \int_r^{\sqrt{4-r^2}} \, r\, dz\, dr\,d\theta

Notice that if we compute the \theta integration first we find

\displaystyle \text{Vol}(S) = 2\pi \, \int_0^{\sqrt{2}}\, \int_r^{\sqrt{4-r^2}} \, r\, dz\, dr = 2\pi \int_0^{\sqrt{2}} r\,\left(\sqrt{4-r^2} - r\right)\,dr

and this integral equals

= 2\pi\int_0^{\sqrt{2}} r\sqrt{4-r^2}\,dr - 2\pi\int_0^{\sqrt{2}} r^2\, dr.

The first of these integrals can be evaluated by applying a u-substitution of u = 4-r^2.

(Spherical Coordinates).  This integral is probably easier to do in spherical coordinates.  Indeed, the region of integration is easily described by the following inequalities:

\displaystyle 0 \leq \rho \leq 2

\displaystyle 0 \leq \theta \leq 2\pi

\displaystyle 0 \leq \varphi \leq \pi/4

The change-of-variables formula gives us

\displaystyle \text{Vol}(S) = \iiint_S \! dV = \int_0^{\pi/4} \, \int_0^{2\pi} \, \int_0^2 \, \rho^2\sin\varphi\,d\rho\,d\theta\,d\varphi

which can be computed relatively easily.

Problem 2.  (a) F(1,0,0) = (a, 0,0), F(0,1,0) = (0, b, 0) and F(0,0,1) = (0, 0, b).  The unit sphere is sent to the ellipsoid whose equation is

\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = \frac{z^2}{c^2} = 1.

In particular, F stretches the x-axis by a factor of a > 0, it stretches the y-axis by a factor of b>0, and it stretches the z-axis by a factor of c > 0.

(b) The derivative matrix of F is easy to compute.  One finds

\displaystyle DF = \left[ \begin{array}{ccc} a & 0 & 0 \\ 0 & b & 0 \\ o & o & c \end{array}\right].

The determinant of this matrix is \det DF = abc.

(c ) To compute the volume of E we can change variables using the function F.  Based on our observations in part (a) we know that if S is the unit sphere, then F(S) = E and so we can convert a volume integral over E as follows:

\displaystyle \text{Vol}(E) = \iiint_E \, dz\,dy\,dx = \iiint_{F(S)} \, dz\,dy\,dx = \iiint_S \, \left|\det\,DF\right|\,du\,dv\,dw\, = abc\iiint_S \, du\, dv\, dw.

This last triple integral computes the volume bounded by the unit sphere S, and we computed that to be 4\pi/3.  Therefore, the volume bounded by the ellipsoid E is given by

\displaystyle \text{Vol} = \frac{4\pi}{3}\,a\cdot b\cdot c.

Problem 3.  This was done in class.

Problem 4.  (a) The length of \gamma is given by

\displaystyle \text{L}(\gamma) = \int_{\gamma} \, ds = \int_0^{2\pi} \, \left|\vec{r}'(t)\right|\,dt = \int_0^{2\pi} \, \sqrt{\sin^2t+\cos^2t + 1} = \int_0^{2\pi} \, \sqrt{2}\, dt = 2\pi\sqrt{2}.

(b) In this problem, the total mass of \gamma is given by the line integral \int_{\gamma} \, f\,ds.  This can be evaluated directly:

\displaystyle \int_{\gamma} \, f\, ds = \int_0^{2\pi} \, f(\vec{r}(t))\,\left|\vec{r}'(t)\right|\,dt = \sqrt{2}\int_0^{2\pi} \cos^2t - \sin^2t + t^2\, dt.

This integral equals \displaystyle \sqrt{2}\,\frac{8\pi^3}{3}.

(c ) The work done by the given vector field (along \gamma) is given by

\displaystyle \int_{\gamma} F\cdot d\vec{s} = \int_0^{2\pi} \, F(\vec{r}(t))\cdot\vec{r}'(t)\,dt.

One can compute the dot product in the integrand as follows:

\displaystyle F(\vec{r}(t))\cdot\vec{r}'(t) = (-\sin t, \cos t, t)\cdot (-\sin t, \cos t, 1) = 1+t.

Hence, the total work done is

\displaystyle \int_0^{2\pi} \, 1+t\, dt = \left[t + \frac{t^2}{2}\right]_0^{2\pi} = 2\pi + 2\pi^2.

Problem 5.  This follows by applying the definition of gradient and curl.  In particular, given such a function f(x,y,z) we have that \nabla f = \left(f_x, f_y, f_z\right).  The curl of the gradient is then given by

\displaystyle \nabla \times \nabla f = \left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \partial_x & \partial_y & \partial_z \\ f_x & f_y & f_z \end{array}\right|.

The resulting vector field is given by

\displaystyle = \left(f_{zy} - f_{yz}\right)\vec{i} - \left(f_{xz} - f_{zx}\right)\vec{j} + \left(f_{xy} - f_{yx}\right)\vec{k} = (0, 0, 0)

since mixed partials are equal.

Problem 6. (a) This line integral can be computed directly, or one can notice that F = \nabla f where f(x,y) = e^{xy}.  By the Gradient Theorem, we then have

\displaystyle \int_{\gamma} F\cdot d\vec{s} = f(4,0) - f(4,0) = 0.

(b) We can apply the same theorem, albeit now to the slightly different curve \alpha, whose start- and end-points are (4,0) and (-4,0), respectively.  We find

\displaystyle \int_{\alpha} F\cdot d\vec{s} = f(-4, 0) - f(4,0) = e^0 - e^0 = 0.

Problem 7.  Here are the statements.

Gradient Theorem.  If F = \nabla f, then

\displaystyle \int_{\gamma} F\cdot d\vec{s} = f(\text{end point}) - f(\text{start point})

Green’s Theorem.  If D is a region in the plane with a correctly oriented boundary, \partial D, then

\displaystyle \iint_D \, \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\, dA = \int_{\partial D}\, \left(P, Q\right)\cdot d\vec{s}

note: if \partial D is parameterized by a function \vec{r}(t) = (x(t), y(t)), then some textbooks will notate d\vec{s} = (x'(t), y'(t))dt and rewrite x'(t) dt = dx and y'(t) dt = dy.  Given this notation, the line integral can also be written as

\displaystyle \int_{\partial D} Pdx + Qdy.

Stokes’ Theorem.  If S is an orientable surface with boundary curve(s) \partial S, then

\displaystyle \iint_S \, \left(\nabla \times F\right) \cdot d\vec{S} = \int_{\partial S} \, F \cdot d\vec{s}.

Here, the line integral is over the oriented curve(s) \partial S whose directions are compatible with the choice of unit normal for the surface S.

Divergence Theorem.  If W is a solid region in space with boundary surface \partial W, then

\displaystyle \iiint_W \, \text{div} F\, dV = \iint_{\partial W} \, F \cdot d\vec{S}

where the surface integral is over \partial W with an outward-pointing normal.

Problem 8.  The area of D is given by

\displaystyle \text{Area}(D) = \iint_D \, dA.

If we can find a vector field F(x,y) = (P(x,y), Q(x,y)) where Q_x - P_y = 1 then we can apply Green’s Theorem to find

\displaystyle \text{Area}(D) = \iint_D \, dA = \int_{\partial D} \, F\cdot d\vec{s}.

There are many, many such vector fields F to choose from.  One choice is to use P(x,y) = 0 and Q(x,y) = 1.  We can then compute the line integral by parameterizing the ellipse-region using \vec{r}(t) = (2\cos t, 5\sin t) for 0 \leq t \leq 2\pi.  (Note: this parameterization gives the boundary curve the correct direction.)  We then have

\displaystyle \text{Area}(D) = \int_0^{2\pi} \, F(\vec{r}(t))\cdot\vec{r}'(t)\,dt = \int_0^{2\pi} 0\cdot (-2\sin t) + (5\cos t)\cdot(5\cos t)\,dt

This integral equals

\displaystyle = \int_0^{2\pi}\, 25\cos^2(t)\, dt = 25\pi

which can be computed by using a double-angle identity.

Problem 9.  In general, the surface area for a surface S is given by

\displaystyle \text{Surface Area}(S) = \iint_S \, dS = \iint_D \, \left|\vec{r}_s\times\vec{r}_t\right|\,dA

where \vec{r}(s,t) : D \to \mathbb{R}^3 parameterizes S.  We can parameterize a graph-surface, S = \left\{(x,y,f(x,y)) : (x,y) \in D\right\} by simply using

\displaystyle \vec{r}(s,t) = (s, t, f(s, t)).

In other words, we are using x = s and y = t.  As worked out in our textbook in 11.6 (on page 231), this becomes

\displaystyle \iint_D \, \sqrt{f_s^2 + f_t^2 + 1} \, ds\, dt

or, if we change notation as is done in the book and use x = s and y = t, then we have

\displaystyle \text{Surface Area}(S) = \iint_D \, \sqrt{f_x^2 + f_y^2 + 1}\, dx\, dy.

For our function we have that f(x,y) = x^2+y^2 and so f_x = 2x and f_y = 2y.  The surface area is then given by

\displaystyle \text{Surface Area}(S) = \iint_D \, \sqrt{4x^2+4y^2 + 1}\, dx\, dy = \int_{-2}^2 \, \int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}\, \sqrt{4x^2+4y^2+1}\,dx\,dy.

Problem 10.  (a) The symbols translate as follows:

ds = \left|\vec{r}'(t)\right|\, dt, \,\, d\vec{s} = \vec{r}'(t)\, dt

dS = \left|\vec{r}_s \times \vec{r}_t\right|\,ds\,dt, \, \, d\vec{S} = \left(\vec{r}_s\times\vec{r}_t\right)\,ds,dt.

(b) The first two are used for line integrals, and the last two are used for surface integrals.

(c ) Again, the surface area of a parameterized S is given by

\displaystyle \text{Surface Area}(S) = \iint_S \, dS = \iint_D \, \left|\vec{r}_s\times\vec{r}_t\right|\,ds\,dt.

A unit normal is given by

\displaystyle \vec{n} = \frac{\vec{r}_s\times\vec{r}_t}{\left|\vec{r}_s\times\vec{r}_t\right|}.

Problem 11.  To compute the flux of F out of S, we need an outward pointing (unit) normal for S.  This requires us to first parameterize the unit sphere S, which we can do as follows:

\displaystyle \vec{r}(s, t) = \left(\sin t\, \cos s, \, \sin t\,\sin s, \, \cos t\right)

where s \in [0, 2\pi] and t \in [0, \pi].  This parameterization comes from spherical coordinates, thinking of s = \theta and t = \varphi.  When we are on the unit sphere, the only restriction (on spherical coordinates) is the equation \rho = 1, leaving \theta and \varphi free to roam within their respective intervals.  If we then recall our conversion formulas relating x, y, z to \rho, \theta, \varphi and set \rho = 1, we obtain the above expression.

A unit normal for S can then be computed by evaluating

\displaystyle \vec{n} = \pm \frac{\vec{r}_s\times\vec{r}_t}{\left|\vec{r}_s\times\vec{r}_t\right|}.

The computation above requires some time to do, but when it is all said and done (and terms are cancelled), we find

\displaystyle \vec{n} = \pm (\sin t\,\cos s, \, \sin t\,\sin s, \, \cos t).

An outward normal is obtained by using the plus sign in the equation above.  This gives us a normal, for example, that points up, out of the sphere, at the north pole (0, 0, 1).  We also find

\displaystyle dS = \left|\vec{r}_s\times\vec{r}_t\right| = \sin t\,ds\,dt.

We can then evaluate this flux as

\displaystyle \iint_S \, F\cdot d\vec{S} = \iint_S \left(F\cdot\vec{n}\right)\,dS = \int_0^{\pi}\,\int_0^{2\pi}\, \sin t\, ds\, dt =4\pi

The integrand F\cdot\vec{n} cancels beautifully; in fact, it equals the constant 1.

(Slightly different approach) One can also approach this problem by simply writing

\displaystyle d\vec{S} = (\vec{r}_s\times\vec{r}_t)\,ds\,dt = \sin t\,(\sin t\,\cos s, \, \sin t\,\sin s, \, \cos t)\,ds\,dt

and then writing out

\displaystyle F(\vec{r}(s,t)) = (x(s, t), y(s, t), z(s, t)) = (\sin t \, \cos s, \, \sin t \, \sin s, \, \cos t).

One can use these expressions to explicitly compute the flux as

\displaystyle \iint_S \, F\cdot d\vec{S} = \int_0^{\pi} \, \int_0^{2\pi} \, F(\vec{r}(s, t))\cdot\left(r_s\times r_t\right)\,ds\,dt.

(Another approach)  One can also observe that, for the unit sphere, \vec{n} = (x,y,z).  Note that this expression for the unit normal does not come from a parameterization, but instead comes from viewing the unit sphere as a level set

\displaystyle x^2+y^2+z^2 = 1.

The gradient of the level-set function G(x,y,z) = x^2+y^2+z^2 is perpendicular to the level set, and so this gradient can be used as a normal.  One finds \nabla G = (2x, 2y, 2z).  To make this vector have unit length we divide by 2 to get (x, y, z) (which has unit length since the point lies on the unit sphere).

Observe that the outward normal for the sphere and the vector field, F, in this problem are the same!  We then have

F\cdot n = x^2+y^2+z^2 = 1

when (x,y,z) lies on the sphere.  This implies that the flux is given by

\displaystyle \iint_S \, F\cdot\vec{n}\,dS = \iint_S \, dS = \text{Surface Area}(S).

This method still leaves one with computing the surface area of the unit sphere, which is probably best done using a parameterization (although some might recall the formula).

(Crazy Cool Approach)  We computed the volume of the solid region bounded by the unit sphere in a previous class and found this volume to be 4\pi/3.  Note that the given sphere, S, is the boundary of the filled-in, solid ball, B = \{(x,y,z) : x^2+y^2+z^2 \leq 1\}, and so we may use the Divergence Theorem to compute the flux:

\displaystyle \iint_S \, F\cdot d\vec{S} = \iint_{\partial B} \, F \cdot d\vec{S} = \iiint_B \, \text{div}(F)\,dV = 3\iiint_B \, dV = 3\cdot \frac{4\pi}{3} = 4\pi.

Problem 12.  (a) To compute this flux directly, we need to first parameterize the surface.  This can be done by using the function

\displaystyle \vec{r}(s, t) = (\cos s, \sin s, t)

where s \in [0, 2\pi] and -1 \leq t \leq 1.  This parameterization is motivated by cylindrical coordinates (with s = \theta, t = z and r = 1 since it is a radius 1 cylinder).

We then compute

\displaystyle d\vec{S} = \left(\vec{r}_s\times\vec{r}_t\right)\,ds\,dt = (-\sin s, \cos s, 0) \times (0, 0, 1) \, ds\, dt = (\cos s, \sin s, 0)\, ds\, dt.

The flux integral is then given by

\displaystyle \iint_S \, F\cdot d\vec{S} = \int_{-1}^1 \, \int_0^{2\pi} \, \big{(}\,(-\sin s, \cos s, t)\cdot(\cos s, \sin s, 0)\big{)}\,ds\,dt = \int_{-1}^1\,\int_0^{2\pi}\,0\,ds\,dt

and so equals zero.

Note: a student might try to do this problem by using, say, the divergence theorem.  After all, the divergence of this vector field is easy to compute, but this is not applicable since the cylinder S is not the boundary of a solid region!  If it contained the “top lid” and “bottom lid,” then it would be.

A student might also try to do this problem by using, say, Stokes’ Theorem.  To do this, though, one would need to know that \nabla \times V = F for some vector field V — however, it is hard to find such a vector field V, and, more over, it does not exist since \text{div}(F) \neq 0.

(b) S is not the boundary of a solid region.  A picture explains this.

(c ) The boundary of S consists of two curves, \gamma_1 and \gamma_2.  Each curve is a circle, one contained in the z = 1 plane and one contained in the z = -1 plane.

The curve \gamma_1 in the z = -1 plane must be oriented in a counter-clockwise direction to be consistent with an outward normal for S.  The curve \gamma_2 in the z = 1 plane must be oriented in a clockwise direction, though!

(d) For this problem one can compute the curl \nabla \times F directly and compute this flux integral directly.  This is a valid way to do this problem since the curl is not that hard to compute.

However, integrating \nabla \times F begs us to use Stokes Theorem.  It says that

\displaystyle \iint_S \left(\nabla \times F\right)\cdot d\vec{S} = \int_{\gamma_1} F\cdot d\vec{s} + \int_{\gamma_2} F\cdot d\vec{s}.

One can parameterize \gamma_1 using the function \vec{r}_1(t) = (\cos t, \sin t, -1) and one can parameterize \gamma_2 using the function \vec{r}_2(t) = (\cos t, \sin t, 1).  Both parameterizations have the domain t \in [0, 2\pi], but the function \vec{r}_2(t) traverses \gamma_2 in the wrong direction, and so we must adjust our line integral with a negative sign.

Both line integrals equal 2\pi, and so when we subtract we (of course) find that

\displaystyle \iint_S \, F\cdot d\vec{S} = 0.

Problem 13.  We can use the divergence theorem to kill this problem.  Observe that the divergence of F is given by \text{div}(F) = 3 and so

\displaystyle 30 = 3\cdot \text{Vol}(W) = 3\cdot \iiint_W \, dV = \iiint_W \, 3dV = \iiint_W \, \text{div}(F)\,dV

and by the Divergence Theorem this last integral equals

\displaystyle \iiint_W \, \text{div}(F)\,dV= \iint_{\partial W} \, F\cdot d\vec{S}.

Therefore the flux out of \partial W = S is 30.


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