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seen Apr 30 '13 at 16:02

Student currently in 2nd year engineering


Apr
29
comment show that a cone is a ruled surface
thank you for the help
Apr
29
comment show that a cone is a ruled surface
Never mind I figured it out. Thanks for the help.
Apr
29
asked Show that product of x, y, and z intercents of tangent plane to surface xyz=1 is a constant
Apr
29
comment show that a cone is a ruled surface
I appreciate the detailed answer but I am still confused. We didn't cover this topic in the course and I just wrote the practice final and again another question about this topic was on it. This time the question was: The hyperboloid of one sheet $x^2+y^2-z^2 = 1$ is called a ruled surface, which is to say that it is made up of straight lines. Prove this by showing that for any $\theta$, the line $$\frac{x - cos\theta}{sin\theta} = \frac{y-sin\theta}{-cos\theta} = z/1$$ lies entirely on the surface... I would really appreciate more help if you don't mind.
Apr
29
comment Find integral of a polar function $h(r,\theta)$ over a circle
The question I originally asked was copied word for word from our review sheet.
Apr
28
comment Find integral of a polar function $h(r,\theta)$ over a circle
So would the integral then be: $\int_0^{2\pi} \int_0^{2acos\theta} rcos\theta rdrd\theta$ ??
Apr
28
asked Find integral of a polar function $h(r,\theta)$ over a circle
Apr
27
revised show that a cone is a ruled surface
edited body
Apr
27
asked show that a cone is a ruled surface
Apr
26
asked How to find points in space where the gradient vector is parallel to another vector
Apr
25
comment tricky surface integral
How did you go from the first line of your integral to the second line with the substitution $x=\frac{1}{\sqrt{2}} tan(t)$. I did the substitution but it brought me to the integral $\sqrt{2} \int_0^1 sec(t) dx$
Apr
25
comment tricky surface integral
Ok. I was also looking online... Could I also use the formula for $\int \sqrt{a^2 +x^2} dx = \frac x2 \sqrt{a^2+x^2} + \frac{a^2}2 ln|x+\sqrt{a^2+x^2}|$ using $a=\sqrt{2}$ and $x=2x$ to get the final answer???
Apr
25
asked tricky surface integral
Apr
24
comment Volume integral over a bounded region
Yes this was helpful however this integral seems to be very complex unless I screwed up. Anyways this is where I learned how to format questions and give detail: meta.math.stackexchange.com/questions/5020/…
Apr
24
comment Volume integral over a bounded region
sorry nvm, I just miss read. you did put the limits of r.
Apr
24
comment Volume integral over a bounded region
I understand this but then what would be the limits of integration for r? Also is it just as correct to calculate the answer using Cartesian coordinates or must it be in spherical?
Apr
24
comment Volume integral over a bounded region
Ok, I got it, however this is a very long process calculation using these constraints for r. Is there any other technique that is easier? I also tried something just to see but since $y\le 1-x$ I subbed this into the formula for z and ended up with the boundary $0 \le z \le 2x-2x^2$ which from there I then let r: 0 ->1 and integrated z subbing in polar coordinates. Is this allowed or no? I feel this was probably stupid but I did get a clean answer.
Apr
24
asked Volume integral over a bounded region
Apr
15
comment Stokes theorem problem to find alpha and beta so that I is independent of the choice of S
Ok. I understand this and computed it to be: $$(2 \beta y(z+1) -1)i + j + yk$$ but using stokes theorem shouldn't $$ \int_C F \bullet dr = \int \int_S curl F \bullet NdS = 0$$ I used this to find $\alpha$ but I can't figure out how use this to find $\beta$ ...
Apr
15
asked Stokes theorem problem to find alpha and beta so that I is independent of the choice of S