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.
$$\mbox{div}(\mathbf{F}\times\mathbf{G})=\mathbf{G}\cdot\mbox{curl}(\mathbf{F})-\mathbf{F}\cdot\mbox{curl}(\mathbf{G})$$