MATH2111 Higher Several Variable Calculus (1 Viewer)

InteGrand · May 6, 2017

Re: Several Variable Calculus

leehuan said:
$\text{Find the surface area of the part of the plane }2x+y +2z = 16 \text{ bounded by the surfaces }\\x = 0, y = 0\text{ and }x^2+y^2=64$

I might be misinterpreting the question but I thought that they're basically asking for 1/4 * area of curve of intersection, which happens to be an ellipse through (0,0,8) passing through (0,8,4) and (8,0,4).

If I'm right, how do I use surface integrals to get to the answer of $24\pi$ ? And if I'm wrong, how do I get back on the right path?

$\noindent Let the surface be $\mathcal{S}$. We know that the area can be expressed using a surface integral as$

$\mathrm{Area}(\mathcal{S}) = \iint _{\mathcal{S}}1 \, \mathrm{d}S.$

$\noindent We just need to figure out how to parametrise $\mathcal{S}$. Here's how to do it. A point $(x,y,z)$ on the surface can be parametrised as $z = \frac{1}{2}\left(12 - y - 2x\right)$, where the parameters used are $x$ and $y$ (i.e. just writing $z$ in terms of $x$ and $y$ from the plane equation, since points on the surface $\mathcal{S}$ are points on the given plane). We just now need to figure out the region in the $x$-$y$ plane that our parameters $x$ and $y$ vary over for our surface. The answer is that they vary over the region $\mathcal{D}:= \left\{(x,y) \in \mathbb{R}^{2} : x^{2} + y^{2} \leq 64, x \geq 0, y \geq 0\right\}$, i.e. the quarter-disk of the disk $x^{2} + y^{2} \leq 64$ that lies in the first quadrant (of the $x$-$y$ plane). Using these facts, you should be able to set up and evaluate the integral.$

leehuan · May 6, 2017

Re: Several Variable Calculus

InteGrand said:
$\noindent Let the surface be $\mathcal{S}$. We know that the area can be expressed using a surface integral as$

$\mathrm{Area}(\mathcal{S}) = \iint _{\mathcal{S}}1 \, \mathrm{d}S.$

$\noindent We just need to figure out how to parametrise $\mathcal{S}$. Here's how to do it. A point $(x,y,z)$ on the surface can be parametrised as $z = \frac{1}{2}\left(12 - y - 2x\right)$, where the parameters used are $x$ and $y$ (i.e. just writing $z$ in terms of $x$ and $y$ from the plane equation, since points on the surface $\mathcal{S}$ are points on the given plane). We just now need to figure out the region in the $x$-$y$ plane that our parameters $x$ and $y$ vary over for our surface. The answer is that they vary over the region $\mathcal{D}:= \left\{(x,y) \in \mathbb{R}^{2} : x^{2} + y^{2} \leq 64, x \geq 0, y \geq 0\right\}$, i.e. the quarter-disk of the disk $x^{2} + y^{2} \leq 64$ that lies in the first quadrant (of the $x$-$y$ plane). Using these facts, you should be able to set up and evaluate the integral.$

Yep makes perfect sense, thanks as always

Out of curiosity, in general does setting x and y to be the parameters make things easier? Or are there scenarios where you'd pick a different choice of parameters.

InteGrand · May 6, 2017

Re: Several Variable Calculus

leehuan said:
Yep makes perfect sense, thanks as always

Out of curiosity, in general does setting x and y to be the parameters make things easier? Or are there scenarios where you'd pick a different choice of parameters.

$\noindent In this one, the surface is one where the points on it have one of the variables (in this case $z$) given explicitly as a function of the other two. In other words, the surface is of the form $z = g(x,y)$, where $x,y$ vary in some region $\mathcal{D}\subseteq \mathbb{R}^{2}$. Therefore, it is natural to just parametrise it as having $z = g(x,y)$, where $x,y$ vary in $\mathcal{D}$. In this case, it is a standard fact (which you can show using the cross product formula for the normal vector) that the normal vector becomes $\mathbf{N} = \left(-g_{x}, -g_{y}, 1\right)$ (where subscripts refer to partial derivatives), so it is easy to get the normal vector too (especially since in this case, the partial derivatives are simple, just constants).$

$\noindent If the surface wasn't one where we had one variable explicitly as a (nice/easy-to-work-with) function of the others, we might parametrise it differently (like by using spherical coordinates or something).$

leehuan · May 6, 2017

Re: Several Variable Calculus

$\text{Find }\int_S \textbf{F}\cdot d\textbf{S}\text{ if }\textbf{F}(\textbf{x})=\phi (||\textbf{x}||)\textbf{x}\text{ is a radial vector field}\\ \text{and }S\text{ is the hemisphere }x^2+y^2+z^2=a^2\text{ with }z\ge 0\\ \text{Hint: use a geometric argument to write down the outward unit normal vector at }S$

InteGrand · May 6, 2017

Re: Several Variable Calculus

leehuan said:
$\text{Find }\int_S \textbf{F}\cdot d\textbf{S}\text{ if }\textbf{F}(\textbf{x})=\phi (||\textbf{x}||)\textbf{x}\text{ is a radial vector field}\\ \text{and }S\text{ is the hemisphere }x^2+y^2+z^2=a^2\text{ with }z\ge 0\\ \text{Hint: use a geometric argument to write down the outward unit normal vector at }S$

$\noindent The outward unit normal at any point $\mathbf{x}$ on the sphere is $\frac{\mathbf{x}}{a}$.$

InteGrand · May 6, 2017

Re: Several Variable Calculus

leehuan said:
$\text{Find }\int_S \textbf{F}\cdot d\textbf{S}\text{ if }\textbf{F}(\textbf{x})=\phi (||\textbf{x}||)\textbf{x}\text{ is a radial vector field}\\ \text{and }S\text{ is the hemisphere }x^2+y^2+z^2=a^2\text{ with }z\ge 0\\ \text{Hint: use a geometric argument to write down the outward unit normal vector at }S$

$\noindent Some further hints. On the sphere $\mathcal{S}$, we have $\mathbf{x}\cdot \mathbf{x} = \left\|\mathbf{x}\right\|^{2} = a^{2}$ and $\left\|\mathbf{x}\right\| = a$, so $\mathbf{F} = \phi(a)\mathbf{x}$ at any point $\mathbf{x}$ on the sphere. Also, $\mathrm{d}\mathbf{S} = \widehat{\mathbf{n}}\, \mathrm{d}S = \frac{\mathbf{x}}{a}\, \mathrm{d}S$, since the outward unit normal at any point $\mathbf{x}$ on the sphere is $\frac{\mathbf{x}}{a}$. Also remember the surface area of a sphere.$

Paradoxica · May 6, 2017

Re: Several Variable Calculus

leehuan said:
$\text{Find the surface area of the part of the plane }2x+y +2z = 16 \text{ bounded by the surfaces }\\x = 0, y = 0\text{ and }x^2+y^2=64$

I might be misinterpreting the question but I thought that they're basically asking for 1/4 * area of curve of intersection, which happens to be an ellipse through (0,0,8) passing through (0,8,4) and (8,0,4).

If I'm right, how do I use surface integrals to get to the answer of $24\pi$ ? And if I'm wrong, how do I get back on the right path?

Obtaining the value of the major radius is slightly tricky, but it's not too bad.

So, the major radius is the hyotenuse of a right-angled triangle running parallel to one of the co-ordinate planes with one of the sides trivially being of length 8.

Now, by parametrising the equation of the circle using trigonometry, it is easy to find the maximum and minimum values of the z-ordinate along boundary of the ellipse, via the method of Auxiliary Transformations (See: Harmonic Addition Theorem). These values turn out to be 8 ± 4√5, which trivially implies the other side length of the triangle is 4√5. By simple computation, the major radius turns out to be 12 units.

Thus, the area of the ellipse is 12 × 8 × π and so the quarter-area that is desired turns out to be 24 π square units.

leehuan · May 12, 2017

Re: Several Variable Calculus

I'm going from cylindrical to spherical in this made-up question.

$\text{The volume of region bound by the cylinder }x^2+y^2=1\text{ and the planes }\\z=\pm 1\text{ is (hopefully)}\\ \int_0^{2\pi}\int_{-1}^1\int_0^1dr\,dz\,d\theta$

$\text{If I try with spherical coordinates instead, would this be right?}\\ \int_0^{2\pi} \int_0^\pi \int_0^{\frac{1}{\sin \phi}} d\rho \,d\phi \, d\theta$

InteGrand · May 12, 2017

Re: Several Variable Calculus

leehuan said:
$\text{If I try with spherical coordinates instead, would this be right?}\\ \int_0^{2\pi} \int_0^\pi \int_0^{\frac{1}{\sin \phi}} d\rho \,d\phi \, d\theta$

$\noindent That would tell us that when $\phi = 0$ for example, $\rho$ is going to go to $\infty$, but our region is bounded, so we should not have $\rho$ going to infinity for this integral.$

leehuan · May 12, 2017

Re: Several Variable Calculus

InteGrand said:
$\noindent That would tell us that when $\phi = 0$ for example, $\rho$ is going to go to $\infty$, but our region is bounded, so we should not have $\rho$ going to infinity for this integral.$

Ah true, that's unfortunate. Forgot that pho follows a different rule for different values of phi. Would I be forced to splitting the integral up into three separate integrals then?

InteGrand · May 12, 2017

Re: Several Variable Calculus

leehuan said:
Ah true, that's unfortunate. Forgot that pho follows a different rule for different values of phi. Would I be forced to splitting the integral up into three separate integrals then?

Yeah, pretty much. Two of those integrals would be equal by symmetry, of course.

leehuan · May 13, 2017

Re: Several Variable Calculus

$\text{For the function }\textbf{f}:\mathbb{R}^2\to \mathbb{R}^2, \, \textbf{f}(x,y) = (e^x\cos y,e^x\sin y)$

$\text{I can understand why }(0,0)\notin \textbf{f}(\mathbb{R}^2)\\ \text{but how would you intuitively realise that it's the only point not in the image of }\textbf{f}?$

InteGrand · May 13, 2017

Re: Several Variable Calculus

leehuan said:
$\text{For the function }\textbf{f}:\mathbb{R}^2\to \mathbb{R}^2, \, \textbf{f}(x,y) = (e^x\cos y,e^x\sin y)$

$\text{I can understand why }(0,0)\notin \textbf{f}(\mathbb{R}^2)\\ \text{but how would you intuitively realise that it's the only point not in the image of }\textbf{f}?$

Imagine it as polar coordinates where the radius is e^x and angle is y ==> radius > 0 (and can attain any positive value) and angle can be anything ==> any point except (0, 0) is in the image (since any nonzero point can be expressed in polar coordinates with a strictly positive radius and some angle).

leehuan · May 14, 2017

Re: Several Variable Calculus

$\int_S \textbf{F}\cdot d\textbf{S}\text{ if }\textbf{F} = \sinh x \textbf{i} + \cosh y \textbf{k}\text{ and if }\\ S\text{ is given by }z=x+y^2\text{ for }0\le y\le x,\, 0\le x \le 1$

I was able to do this question correctly using the typical approach and the answer is sinh(1) - 1. Here's an approach that seems to be wrong.

$\text{Consider the parametrisation }\textbf{r} = \begin{pmatrix}x\\y\\ x+y^2\end{pmatrix}\text{ with the same restrictions}\\ \text{Then, }\frac{d\textbf{r}}{dy} \times \frac{d\textbf{r}}{dx} =\begin{pmatrix}0\\1\\2y\end{pmatrix}\times \begin{pmatrix}1\\0\\1\end{pmatrix} = \begin{pmatrix}1\\ 2y\\ -1\end{pmatrix} \\\text{and }\textbf{F}(\textbf{r}) = \sinh x \textbf{i} + \cosh y \textbf{k}$

$\text{Then }\int_S \textbf{F}\cdot d\textbf{S} = \int_0^1 \int_0^x\textbf{F}(\textbf{r}) \cdot \left[\frac{d\textbf{r}}{dy} \times \frac{d\textbf{r}}{dx} \right] dy\, dx\\=\int_0^1 \int_0^x (\sinh x - \cosh y)\, dy\, dx =\dots = 1-\sinh 1$

Which line is the mistake in and why is it incorrect?

InteGrand · May 14, 2017

Re: Several Variable Calculus

leehuan said:
$\int_S \textbf{F}\cdot d\textbf{S}\text{ if }\textbf{F} = \sinh x \textbf{i} + \cosh y \textbf{k}\text{ and if }\\ S\text{ is given by }z=x+y^2\text{ for }0\le y\le x,\, 0\le x \le 1$

I was able to do this question correctly using the typical approach and the answer is sinh(1) - 1. Here's an approach that seems to be wrong.

$\text{Consider the parametrisation }\textbf{r} = \begin{pmatrix}x\\y\\ x+y^2\end{pmatrix}\text{ with the same restrictions}\\ \text{Then, }\frac{d\textbf{r}}{dy} \times \frac{d\textbf{r}}{dx} =\begin{pmatrix}0\\1\\2y\end{pmatrix}\times \begin{pmatrix}1\\0\\1\end{pmatrix} = \begin{pmatrix}1\\ 2y\\ -1\end{pmatrix} \\\text{and }\textbf{F}(\textbf{r}) = \sinh x \textbf{i} + \cosh y \textbf{k}$

$\text{Then }\int_S \textbf{F}\cdot d\textbf{S} = \int_0^1 \int_0^x\textbf{F}(\textbf{r}) \cdot \left[\frac{d\textbf{r}}{dy} \times \frac{d\textbf{r}}{dx} \right] dy\, dx\\=\int_0^1 \int_0^x (\sinh x - \cosh y)\, dy\, dx =\dots = 1-\sinh 1$

Which line is the mistake in and why is it incorrect?

You ended up with the negative of your previous answer because you used the oppositely oriented normal.

leehuan · May 14, 2017

Re: Several Variable Calculus

InteGrand said:
You ended up with the negative of your previous answer because you used the oppositely oriented normal.

Oh. In general, how do we determine which normal vector has the correct orientation?

leehuan · May 25, 2017

Re: Several Variable Calculus

$\text{Prove (using the definition) that }f_n:[0,1]\to \mathbb{R}\\ f_n(x) = n^\alpha x^n(1-x)\to 0\text{ pointwise}$

$\alpha \in \mathbb{R}\text{ is fixed}$

(Also ignore the previous question, even the lecturer agreed it was lacking information)

InteGrand · May 25, 2017

Re: Several Variable Calculus

leehuan said:
$\text{Prove (using the definition) that }f_n:[0,1]\to \mathbb{R}\\ f_n(x) = n^\alpha x^n(1-x)\to 0\text{ pointwise}$

$\alpha \in \mathbb{R}\text{ is fixed}$

(Also ignore the previous question, even the lecturer agreed it was lacking information)

The basic ideas are:

- When x = 0 or 1 (the endpoints of the domain), f_n (x) = 0 for every n, so the convergence at these x is clear.

- For any given x in (0, 1) (interior of the domain), the fact that exponential decay dominates power function growth yields the result.

leehuan · May 26, 2017

Re: Several Variable Calculus

Got that part out.

$\text{Now how would I prove uniform convergence for }\alpha < 1?$

InteGrand · May 26, 2017

Re: Several Variable Calculus

leehuan said:
Got that part out.

$\text{Now how would I prove uniform convergence for }\alpha < 1?$

What was your progress so far?

MATH2111 Higher Several Variable Calculus (1 Viewer)

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