Mark's Calc III - Individual question feed

Final Review Number one

asmith14 — Thu, 31 Jul 2014 17:38:09 -0500

Let D denote the solid pyramid with vertices located at (4,0,0), (0,1,0), (0,0,2), and the origin. Set up $\iiint_{D} f(x,y,z)dV$. So when we were talking about this in class we drew the graph first and then set up the equation ax+by+cz=d and then d was substituted for 4 because it was the least common multiple of the points. My question is how were the other numbers found in the equation which made it x+4y+2z=4 ?

Partial derivatives

Tiffany — Thu, 31 Jul 2014 11:05:02 -0500

When showing that $\vec F$ is conservative, we take the partial derivatives, just wondering if that is the same as $\nabla f $? I know from there we find the function $f$ to make them equal? But I thought that the gradient was the partial derivatives?

Volume integrals

Tiffany — Thu, 31 Jul 2014 11:01:05 -0500

I was wondering if when we are setting up volume integrals, if we always integrate as 1? then $\delta \rho \delta \phi \delta \theta$ ? or do we use the function, and then if it ends up being 1, we go with that? For example number 5 on the review sheet for the final exam. It says "Let S denote the unit sphere. Evaluate $$\iiint_S (x^2 + y^2 + z^2) dV$$ " So I was wondering if we set it up as... $$ \int _0 ^{2\pi} \int _0 ^\pi \int _0^1 \rho^2* \rho^2 \sin \phi \delta \rho \delta \phi \delta \theta $$ or $$ \int _0 ^ {2\pi} \int _0 ^\pi \int _0^1 1 * \rho^2 \sin \phi \delta \rho \delta \phi \delta \theta $$

Final Review

Tiffany — Wed, 30 Jul 2014 20:48:12 -0500

I feel like problem number 2 on the exam review should be easy, and I'm just making it hard for no reason. The problem states "Find the volume in the first octant (x, y, z all positive) and under the graph of the function $f(x,y)=4-(x^2 + y^2).$" With it being in the first octant, $\delta \theta$ should go from 0 to $2\pi$ since we need 1/4 of a circle. And r would go from 0 to 4? This part i'm not exactly sure of, I know that $(x^2 + y^2)$ is a circle, I'm not exactly sure what the 4- goes to, and graphing it on wolfram alpha didn't help much. It looks like it extends it up to 4 in the z direction? So this may be where my original mistake is. So i'm going to work this two ways. The first with $0\leq r \leq 4$ and the second being $ 0\leq r \leq 1$ With my reasoning behind one, because of the $x^2 + y^2$ denoting the unit circle. $$ \int _0 ^{\frac \pi 2} \int _0 ^4 4-r^2 r \delta r \delta \theta $$ $$ = \frac \pi 2 \int _0 ^4 4-r^3 \delta r$$ $$ = \frac \pi 2 * 4r- \frac 14 r^4 \bigg |_0 ^4 $$ $$ = \frac \pi 2 * 16-64$$ $$ = \frac \pi 2 * -48 $$ $$= \frac {-48\pi}2$$ $$ = -24\pi$$ and the second being $$ \int _0 ^{\frac \pi 2} \int _0 ^1 4-r^2 r \delta r \delta \theta $$ $$ = \frac \pi 2 \int _0 ^1 4-r^3 \delta r$$ $$ = \frac \pi 2 * 4r- \frac 14 r^4 \bigg |_0 ^1 $$ $$ = \frac \pi 2 * 4-\frac 14$$ $$ = \frac \pi 2 * \frac{15}4 $$ $$= \frac {15\pi}8$$ I'm not sure if either of these are right. So if not, if someone could explain the easiest way to think about this?

Review for final Exam

asmith14 — Tue, 29 Jul 2014 18:56:25 -0500

I was looking back over Exam 1 and question number ten is still giving me a problem. Can someone please explain the right path that I should have taken to get the right answer? The question was Let $p(t)=<1+3t, 2+4t>$ and let $q(t)=<3-t, -4+3t>$. Find the point of intersection between the paths parametrized by p and q. What I attempted was to solve for u and t $$1=3t+3-u \rightarrow -2+3t=-u \rightarrow (2-3t)=u$$ $$2+4t=-4-3u$$ $$2+4t=-4-3(2-3t)$$ $$6+4t=-6+9t$$ $$12=5t$$ $$t=12/5$$ but then I don't know what to do complete the problem after that.