# Thread: MATH2111 Higher Several Variable Calculus

1. ## Re: Several Variable Calculus

Originally Posted by leehuan
Just a brief sketch-out please

$f(x)=\begin{cases}0 & (x,y)=(0,0)\\ \frac{2xy}{x^2+y^2} & \text{otherwise}\end{cases}$

$\text{Why is it that }\frac{\partial^2 f}{\partial x \partial y}(0,0)\text{ DNE?}$

I know that f is not continuous at 0 but I'm not sure if that helps since we're talking about partial derivatives here.
If we switch to polar co-ordinates, then the result is:

$\frac{\partial^2 f}{\partial x \partial y} = \frac{-2\cos{4\theta}}{r^6}$

It is then immediately obvious that the limit does not exist as r tends towards zero.

2. ## Re: Several Variable Calculus

Originally Posted by leehuan
Just a brief sketch-out please

$f(x)=\begin{cases}0 & (x,y)=(0,0)\\ \frac{2xy}{x^2+y^2} & \text{otherwise}\end{cases}$

$\text{Why is it that }\frac{\partial^2 f}{\partial x \partial y}(0,0)\text{ DNE?}$

I know that f is not continuous at 0 but I'm not sure if that helps since we're talking about partial derivatives here.
You can try first computing an expression for ∂f/∂y (using first principles to find the value of this at the origin). Then try computing ∂2f/∂x∂y (i.e. ∂/∂x (∂f/∂y)) at the origin from first principles and show that the limit does not exist.

3. ## Re: Several Variable Calculus

Oh damn, looks like the brute way is the only way out of that one. Was hoping for a shortcut
_________________________

I got asked a question but I haven't seen the notation before

$\text{Let }\phi =-4xy^3+z^2x$

$\text{Compute }\nabla \phi\text{ and verify that }\nabla \times (\nabla \phi) = \textbf{0}$

What is that isolated nabla supposed to mean?

4. ## Re: Several Variable Calculus

Nabla phi I am sure you understand, that is the gradient of phi and is a vector-valued function. Which can be regarded as a triple of functions (f,g,h) from R^3 to R.

Nabla cross (f,g,h) denotes the curl of the vector-valued function (f,g,h).

If you write Nabla as (d/dx,d/dy,d/dz), then Nabla x (f,g,h) is defined the same way as the usual cross product between two 3-d vectors. (Just the first "vector" consists of a triple of differential operators, whilst the second consists of a triple of real-valued functions on R^3. So instead of multiplying real numbers (vector components) together as in the defn of the usual cross product, we are applying differential operators to functions.)

eg first component is dh/dy-dg/dz.

5. ## Re: Several Variable Calculus

(The notation Nabla dot (f,g,h) is the divergence of a vector field and is understood in the same way.)

6. ## Re: Several Variable Calculus

If you keep cheating you won't learn. Why are marks so important?

7. ## Re: Several Variable Calculus

Seriously? If I am stuck on a question why can I not ask for help? Accusing me of cheating on an assignment if I genuinely can't do my homework is just low.

8. ## Re: Several Variable Calculus

So for a continuous mapping f: are the following always true

(I didn't really define the domain and codomain of f so just assume whatever's convenient please)

1) U is closed => f^-1(U) is closed
2) U is open => f^-1(U) is open
3) U is closed => f(U) is closed
4) U is open = f(U) is open

5) U is compact => f(U) is compact
6) U is path-connected => f(U) is path connected

Don't really need proof, just yes/no is sufficient

9. ## Re: Several Variable Calculus

Originally Posted by leehuan
So for a continuous mapping f: are the following always true

(I didn't really define the domain and codomain of f so just assume whatever's convenient please)

1) U is closed => f^-1(U) is closed
2) U is open => f^-1(U) is open
3) U is closed => f(U) is closed
4) U is open = f(U) is open

5) U is compact => f(U) is compact
6) U is path-connected => f(U) is path connected

Don't really need proof, just yes/no is sufficient
1) True
2) True
3) False
4) False
5) True
6) True

10. ## Re: Several Variable Calculus

There was a question in my test that I could not do and I'm seeking a solution please

$\text{Determine with explanations whether or not the point }\textbf{a}=(0,1)\\ \text{is a boundary point for the set}\\ S=\{(x,y)\in \mathbb{R}^2 : x\neq 0, y=\sin \frac1x \}$

11. ## Re: Several Variable Calculus

Originally Posted by leehuan
There was a question in my test that I could not do and I'm seeking a solution please

$\text{Determine with explanations whether or not the point }\textbf{a}=(0,1)\\ \text{is a boundary point for the set}\\ S=\{(x,y)\in \mathbb{R}^2 : x\neq 0, y=\sin \frac1x \}$
Yeah it is a boundary point. Here's some hints of a possible method. Note that S is just the graph of y = f(x) := sin(1/x) (x =/= 0). Use (or show) the fact that f attains the value 1 for arbitrarily small values of x > 0 to deduce that (0, 1) is a boundary point of S (noting that any ball around (0, 1) contains points not in S also, as there will be points in it with y-value greater than 1).

12. ## Re: Several Variable Calculus

I get lost in this Einstein summation convention so can I see how this problem would be approached using it?

$\text{For a general }\textbf{a}=(a_1,a_2,a_3)^T \in \mathbb{R}^3\text{ find the matrix }A\in M_{3,3}(\mathbb{R})\\ \text{such that }\textbf{a}\times \textbf{v}=A\textbf{v}\\\forall \textbf{v}\in \mathbb{R}^3$

13. ## Re: Several Variable Calculus

Originally Posted by leehuan
I get lost in this Einstein summation convention so can I see how this problem would be approached using it?

$\text{For a general }\textbf{a}=(a_1,a_2,a_3)^T \in \mathbb{R}^3\text{ find the matrix }A\in M_{3,3}(\mathbb{R})\\ \text{such that }\textbf{a}\times \textbf{v}=A\textbf{v}\\\forall \textbf{v}\in \mathbb{R}^3$
$\noindent For r\in\left\{1,2,3\right\}, the r-th column of A (which will be a 3\times 3 matrix) will be equal to A\mathbf{e}_{r} = \mathbf{a}\times \mathbf{e}_{r}. So try and find an expression for \mathbf{a}\times \mathbf{e}_{r} using the Einstein summation convention, using the formula for a cross product and noting that the j-th element of \mathbf{e}_{r} is \delta_{rj}.$

14. ## Re: Several Variable Calculus

$\text{Express the volume above the cone }z=\sqrt{x^2+y^2}\\\text{and inside the sphere }x^2+y^2+z^2=2az\\ \text{using Cartesian, cylindrical and spherical coordinates}$

Even with the diagram in front of me I still struggle to figure out my boundaries of integration.

$\text{'Progress' with Cartesian coordinates (unsure if right path)}\\ V=\int_{-a}^a \int_{?}^{?} \int_{x}^{a+\sqrt{a^2-x^2-y^2}} dz\,dy\,dx$

15. ## Re: Several Variable Calculus

Originally Posted by leehuan
$\text{Express the volume above the cone }z=\sqrt{x^2+y^2}\\\text{and inside the sphere }x^2+y^2+z^2=2az\\ \text{using Cartesian, cylindrical and spherical coordinates}$

Even with the diagram in front of me I still struggle to figure out my boundaries of integration.

$\text{'Progress' with Cartesian coordinates (unsure if right path)}\\ V=\int_{-a}^a \int_?^? \int_{x}^{a+\sqrt{a^2-x^2-y^2}}dz\,dy\,dx$
$\noindent Note that this is an ice-cream cone'' shape where the sphere intersects the cone above the x-y plane in the horizontal plane z = a (assuming a > 0) and makes a circle with x^{2} + y^{2} = a^{2} in this plane in the intersection. So if we cast the shadow of the solid down onto the x-y plane, it is the disk x^{2} + y^{2} \leq a^{2}. So the iterated integral can have x and y vary over the disk \mathcal{D} := \left\{(x,y) \in \mathbb{R}^{2}: x^{2} + y^{2} \leq a^{2}\right\} and then for each fixed (x,y) in this disk \mathcal{D}, we have z go from the cone to the spherical cap (which is just the upper hemisphere, where z = a + \sqrt{a^{2} - x^{2} - y^{2}}, which we get from solving the sphere's equation for z and taking the root with z \geq a), i.e. from \sqrt{x^{2} + y^{2}} to a + \sqrt{a^{2} - x^{2} - y^{2}}. Hence the integral in Cartesian coordinates becomes$

\begin{align*}\iint_{\mathcal{D}} \int_{z = \sqrt{x^{2} + y^{2}}}^{z = a + \sqrt{a^{2} - x^{2} - y^{2}}} \mathrm{d}z \, \mathrm{d}y \, \mathrm{d}x = \int_{x = -a}^{x=a} \int_{y = -\sqrt{a^{2} - x^{2}}}^{y= \sqrt{a^{2} - x^{2}}} \int_{z = \sqrt{x^{2} + y^{2}}}^{z = a + \sqrt{a^{2} - x^{2} - y^{2}}} \mathrm{d}z \, \mathrm{d}y \, \mathrm{d}x.\end{align*}

16. ## Re: Several Variable Calculus

I think spherical is preferred here (could be mistaken). But using the best approach, how would you determine the limits for this

$\text{Find the volume of the solid bounded between the spheres }\\x^2+y^2+z^2=a^2\, \text{and }x^2+y^2+(z-a)^2 = a^2$

17. ## Re: Several Variable Calculus

Originally Posted by leehuan
I think spherical is preferred here (could be mistaken). But using the best approach, how would you determine the limits for this

$\text{Find the volume of the solid bounded between the spheres }\\x^2+y^2+z^2=a^2\, \text{and }x^2+y^2+(z-a)^2 = a^2$
Inspection would be the best way to determine the limits probably (you can draw a diagram to help if you decide to use inspection).

18. ## Re: Several Variable Calculus

Originally Posted by InteGrand
Inspection would be the best way to determine the limits probably (you can draw a diagram to help if you decide to use inspection).
I was able to deduce that $0\le \theta\le 2\pi\text{ and }0\le \phi \le \frac{\pi}{2}$

It's always rho that gets me.

Unless I picked the order of integration wrong.

19. ## Re: Several Variable Calculus

Originally Posted by leehuan
I was able to deduce that $0\le \theta\le 2\pi\text{ and }0\le \phi \le \frac{\pi}{2}$

It's always rho that gets me.

Unless I picked the order of integration wrong.
It might be easier to use cylindrical coordinates. The way rho varies is different depending on what range phi is in. When phi is such that your point in the region lies within the "ice cream cone" part of the region, then rho will vary from 0 to a. For phi larger than the angle make by the cone to the vertical, rho will vary from 0 to the value of rho at the point on the sphere x^2 + y^2 + (z-a)^2 = a^2 with this phi value. You can find these using the relations between Cartesian and Spherical Coordinates.

20. ## Re: Several Variable Calculus

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

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?

21. ## Re: Several Variable Calculus

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

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

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

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

22. ## Re: Several Variable Calculus

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

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

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

23. ## Re: Several Variable Calculus

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

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

24. ## Re: Several Variable Calculus

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

25. ## Re: Several Variable Calculus

Originally Posted by leehuan
$\text{Find }\int_S \textbf{F}\cdot d\textbf{S}\text{ if }\textbf{F}(\textbf{x})=\phi (||\textbf{x}||)\textbf{x}\text{ is a radial vector field}\\ \text{and }S\text{ is the hemisphere }x^2+y^2+z^2=a^2\text{ with }z\ge 0\\ \text{Hint: use a geometric argument to write down the outward unit normal vector at }S$
$\noindent The outward unit normal at any point \mathbf{x} on the sphere is \frac{\mathbf{x}}{a}.$

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