Practice Problems for April 27 2011

1. Evaluate the surface integral

$$\iint_S z~dS,$$

where $S$ is part of the paraboloid $z=x^2+y^2$ under the plane $z=9$.

Solution:
Using the tools from class we can write this integral as follows.
$$\iint_S z~dS=\iint_D(x^2+y^2)\sqrt{1+4x^2+4y^2}dA$$
Where $D$ is the disk $x^2+y^2\leq9$. After changing to polar coordinates we have
$$\int_0^{2\pi}\int_0^3r^3\sqrt{1+4r^2}drd\theta=2\pi\int_0^3r^3\sqrt{1+4r^2}dr,$$
which may be evaluated using trigonometric substitution. Let $r=\frac{1}{2}\tan\phi$, so $dr=\frac{1}{2}\sec^2\phi$ and $\sqrt{1+4r^2}=\sqrt{\sec^2\phi}=\sec\phi$. Thus the integral is transformed into
$$\int\frac{1}{16}\tan^3\phi\sec^3\phi d\phi=\frac{1}{16}\int \tan\phi(\sec^2\phi-1)\sec^3\phi d\phi$$
$$=\frac{1}{16}\int (\sec^5\phi-\sec^3\phi)\tan\phi\sec\phi d\phi$$
Now use $u$-substitution with $u=\sec\phi$ and thus $du=\tan\phi\sec\phi d\phi$ and so the integral becomes
$$\frac{1}{16}\int(u^5-u^3)du=\frac{1}{96}u^6-\frac{1}{64}u^4$$
reversing the substitutions we have
$$=\frac{1}{96}\sec^6\phi-\frac{1}{64}\sec^4\phi$$.
Then using $\sec\phi=\sqrt{1+\tan^2\phi}$ this becomes
$$=\frac{1}{96}(1+4r^2)^3-\frac{1}{64}(1+4r^2)^2.$$
Now evaluate at $r=3$ and $r=0$ and we are all done.

2. Evaluate the surface integral

$$\iint_S x~dS,$$

where $S$ is the bottom half of the sphere $x^2+y^2+z^2=4$.

Posted by: Angel Castro

Solution

3. Evaluate the surface integral

$$\iint_S (x^2+y^2+z^2)dS,$$

where $S$ is the cylinder $x^2+z^2=1$ between $y=-1$ and $y=1$, together with the top and bottom disks.

Solution: by Dima Tarasevich

4. Suppose $\mathbf{F}(x,y,z)=x\mathbf{i}-z\mathbf{j}+y\mathbf{k}$ and $S$ is the part of the sphere $x^2+y^2+z^2=16$ in the first octant. Find

$$\iint_S \mathbf{F}\cdot d\mathbf{S}.$$

5. Suppose $\mathbf{F}(x,y,z)=xy\mathbf{i}+yz\mathbf{k}$ and $S$ is parameterized by $\mathbf{r}(u,v)=(u+v)\mathbf{i}+uv\mathbf{j}+(u-v)\mathbf{k}$ for $(u,v)\in [0,1]\times[0,1]$. Find

$$\iint_S \mathbf{F}\cdot d\mathbf{S}.$$

SOLUTION:

$\mathbf{r}u X \mathbf{r}v = (-v-u)\mathbf{i}+ 2\mathbf{j} + (u-v)\mathbf{k}$

$\mathbf{F}\cdot \mathbf{r}u X \mathbf{r}v = (u+v)(-v-u) +2uv + (u-v)(u-v)$

$$\int_0^1\int_0^1(u+v)(-v-u)\mathbf{i} +2uv\mathbf{j} + (u-v)(u-v)\mathbf{k}$$

$$\int_0^1\int_0^1 2uv dudv = 1/2$$

Charlie

Practice Problems for April 26 2011

Here is a collection of problems related to what we did today in class. Feel free to add solutions

1. Find the parametric equations of the cylinder $\frac{y^2}{4}+9z^2=1$

2. Find the parametric equations of the sphere $x^2+y^2+z^2=25$ above the cone $z=\sqrt{x^2+y^2}$. What values of the parameters are needed?

3. Find the surface area of the part of the sphere $x^2+y^2+z^2=1$ between $z=-\frac{1}{2}$ and $z=\frac{1}{2}$.

Solution: First we'll parameterize the sphere by $x=\cos\theta\sin\phi$, $y=\sin\theta\sin\phi$, and $z=\cos\phi$. So we can write $\mathbf{r}(\theta,\phi)=\cos\theta\sin\phi\mathbf{i}+\sin\theta\sin\phi\mathbf{j}+\cos\phi\mathbf{k}$. In order to find the values of $\theta$ and $\phi$ that "draw" the needed portion of the sphere we set  $z=\pm \frac{1}{2}$ and so

$$x^2+y^2=\frac{3}{4}$$

after plugging in the parametric equations we are left with

$$\sin^2\phi=\frac{3}{4}$$

and thus

$$\sin\phi=\pm\frac{\sqrt{3}}{2}.$$

This corresponds to $\frac{\pi}{3}\leq\phi\leq\frac{5\pi}{3}$(looking it up here). The values of $\theta$ are from $0$ to $2\pi$. Now we are ready.  Recall the formula for surface area

$$A(S)=\iint_S dS=\iint_D |\mathbf{r}_u\times\mathbf{r}_v|dA,$$

where $D$ are the necessary values of $u$ and $v$ to create $S$. First we'll calculate

$$\mathbf{r}_{\phi}\times\mathbf{r}_\theta=\mbox{Det}\pmatrix{\mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos\theta\cos\phi & \sin\theta\cos\phi & -\sin\phi \\ -\sin\theta\sin\phi & \cos\theta\sin\phi & 0}.$$
Putting it all together
$$A(S)=\int_{0}^{2\pi}\int_{\pi/3}^{5\pi/3}\sin\phi d\phi d\theta$$
$$~~2\pi(\cos(\pi/3)-\cos(5\pi/3))=2\pi$$.

4. Find the area of the boundary of the region enclosed by $z=x^2+y^2$ and $z=9$.

5. Find the area of the surface $z=\sqrt{x^3}+\sqrt{y^3}$ with $(x,y)\in[0,1]\times [0,1]$.

Solution: Recall in this special case we can use
$$A(S)=\int_D \sqrt{1+(\frac{\partial z}{\partial x})^2(\frac{\partial z}{\partial y})^2}dA$$, where
$D=[0,1]\times [0,1]$, $\frac{\partial z}{\partial x}=\frac{3}{2}\sqrt{x}$, and $\frac{\partial z}{\partial y}=\frac{3}{2}\sqrt{y}$. So
$$A(S)=\int_0^1\int_0^1\sqrt{1+\frac{9}{4}x+\frac{9}{4}y}dxdy$$
$$~~~=\int_0^1\frac{8}{27}((\frac{13}{4}+\frac{9}{4}y)^{3/2}-(1+\frac{9}{4}y)^{3/2})dy,$$
and you can take it from here.

6. Find the are of the part of the cylinder $x^2+y^2=4$ between $z=0$ and $z=x+1$.

Exam 2 Corrections and New Examples

As we did last time, I am allowing you to make corrections on the second test. The rules are the same as they were before:

• Corrections will be due this thursday afternoon(April 14). If I'm not around slip it under my door. It will be considered "on time" if its under my door when I arrive on Friday morning.
• Each question was worth 12 points. You can earn up to half of the points you missed for each question.
• I don't want your test book back, but make sure to write how many points you earned for each problem on the sheet you turn in.
• If you didn't keep the exam, you can find it in the download page of this blog.
• The bonus question will become a bonus question in second written homework assignment.

Now for a few examples:

Example 1: Let $\mathbf{F}(x,y)=(2x+ye^{xy})\mathbf{i}+(2y+xe^{xy})\mathbf{j}$ and $C$ be the curve parameterized by $x=te^{t^2-t}+1$,
$y=\sin \frac{\pi}{2}t+\cos \frac{\pi}{2}t$ for $t\in[0,1]$. Find
$$\int_C \mathbf{F}\cdot d\mathbf{r}$$.

Example 2: Let $C$ be the boundary of the triangle with vertices $(0,0)$, $(1,1)$, and $(1,0)$. Find the following integral.

$$\oint_{C} xydx+2x^2y^2 dy$$

-Abigail Walker


Example 3: Let $C$ be the ellipse defined by $\frac{x^2}{4}+\frac{y^2}{9}=1$. Evaluate

$$\oint_C ydx+xydy.$$

Example 4: Evaluate

$$\iiint_{E}\sqrt{x^2+y^2}dV,$$

where $E$ is the region enclosed by the planes $z=0$, $z=x+y+5$, and the
cylinders $x^2+y^2=4$ and $x^2+y^2=9$.


Example 5: Let $\mathbf{F}(x,y,z)=(xyz+2z^2)\mathbf{i}+\cos(xyz)\mathbf{j}+xy^2z^3\mathbf{k}$. Find $\mbox{div}~\mathbf{F}$ and $\mbox{curl}~\mathbf{F}$.

SOLUTION:

-Sabrina Campfield

Example 6: Let $\mathbf{F}(x,y,z)=e^{x^2+y^2+z^2}\mathbf{i}-\arctan(xyz)\mathbf{k}$. Find $\mbox{div}~\mathbf{F}$ and $\mbox{curl}~\mathbf{F}$.

Example 7: If the components of $\mathbf{F}$ have continuous second partial derivatives, verify the identity.
$$\mbox{div}~\mbox{curl}~\mathbf{F}=0$$

Solution:

Damian Miraglia


Example 8: Verify the identity.
$$\mbox{div}(\mathbf{F}\times\mathbf{G})=\mathbf{G}\cdot\mbox{curl}(\mathbf{F})-\mathbf{F}\cdot\mbox{curl}(\mathbf{G})$$

Written Homework 2 Hints

1.  First off, if you look at this problem the right way it turns becomes very similar to one we worked out in class and was on the second test. Here is a similar problem.

$$\int_0^1e^{\mbox{max}\{x,1-x\}}dx$$.

First we'll split up the region of integration into two pieces, one where $x$ is larger and one where $1-x$ is larger. Observe that $1-x$ is larger on the interval $[0,1/2]$ and $x$ is larger on the interval $[1/2,1]$. So we can write

$$\int_0^1e^{\mbox{max}\{x,1-x\}}dx=\int_{0}^{1/2}e^{1-x}dx+\int_{1/2}^1 e^xdx,$$

which are fairly simple.

2.  The hint written on the assignment is sufficient.

3.  Use the formula for 4 dimensional volume(similar to the one for area).

$$V(D)=\iiiint_{D}dV,$$ 

and use change of variables to write 

$$dV=\left|\frac{\partial(x,y,z,w)}{\partial(r,\theta,s,\varphi)}\right|,$$

and use the technique here to take the determinant. Be careful about the bounds of integration for $r$ and $s$.

Thats all for now, we'll have the technology to solve 4 and 5 after this week.

Exam 2 Notes and Examples

Our second exam is coming up day after tomorrow. In light of this I would like to relay a few things that I have said in class.

1. You can present a question from the review sheet in class tomorrow(Wednesday 6 April) to earn 5 bonus points on the exam.
2. You will be allowed a formula sheet that you write. It may be up to one page(only front), and may include no examples.
3. I'm putting some examples you should look at below, feel free to add your solution to fulfill your "online collaboration" credit.

Example 1:

Calculate the following line integral, where $C$ is the left half of a circle with radius 3.

$$ \int_{C}(x^{2}y+xy^2)ds $$

Solution: Taking inspiration from polar coordinates, we will parameterize $C$ by $\mathbf{r}(t)=3\cos t\mathbf{i}+3\sin t\mathbf{j}$, with $t\in[\pi/2,3\pi/2]$. A simple calculation gives

$$ds=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}=3.$$

So the integral becomes

$$\int_{C}(x^{2}y+xy^2)ds=81\int_{\pi/2}^{3\pi/2}(\cos^2t\sin t+\cos t\sin^2t)dt$$

$$=(27\sin^3t)_{\pi/2}^{3\pi/2}=-54.$$

Example 2:

Calculate the line integral, where $C$ is the line segment from $(1,-2)$ to $(4,1)$.

$$\int_{C}(xe^y+y\sin(x))ds$$
Solution:
We can parameterize the line segment using $\mathbf{r}(t)=\left<1,-2\right>(1-t)+\left<4,1\right>t$, so we have $x=1+3t$ and $y=-2+3t$. This gives us $ds=\sqrt{9+9}dt=3\sqrt{2}dt$ and thus
$$\int_{C}(xe^y+y\sin(x))ds=3\sqrt{2}\int_0^1((1+3t)e^{-2+3t}+(-2+3t)\sin(1+3t))dt$$

$$~~~~=[te^{-2+3t}-\frac{1}{3}(2+3t)\cos(1+3t)+\frac{1}{3}\sin(1+3t)]_{t=0}^{t=1}$$
$$~~~~=\mbox{you can get the rest}$$

Example 3:

Let $C$ be the curve made up of the line segment from $(-1,0)$ to $(0,0)$, the curve $y=x^2$ from $(0,0)$ to $(1,1)$, and the line segment from $(1,1)$ to $(1,2)$. Calculate the line integral.

$$\int_{C}xds.$$

Solution

I would suggest working these out before the exam and then if you want to post a solution do so between thursday and tuesday. This way you still have time to study. If there are no solutions by next tuesday I'll post solutions, but I warn you guys there are only going to be a limited number of chances to write these solutions so you might consider "getting it done" soon.