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
$$A(S)=\int_{0}^{2\pi}\int_{\pi/3}^{5\pi/3}\sin\phi d\phi d\theta$$
$$~~2\pi(\cos(\pi/3)-\cos(5\pi/3))=2\pi$$.
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.
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 $z=x+1$.
No comments:
Post a Comment