Mark's Calc III - Individual question feed

Exam III review sheet

Anonymous — Wed, 30 Jul 2014 10:08:58 -0500

So I am completely confused on how to solve #1 on the final exam review sheet: Let $D$ denote the solid pyramid with vertices located at $(4,0,0),(0,1,0),(0,0,2)$ and the origin. Set up $$\iiint_D f(x,y,z)dV$$ in the order $dzdydx$. We did one similar to this in class but we were given an equation for the plane. I guess the real question is, how would we go about finding an equation of a plane that we could work with?

Can Anyone Validate Problem #3

Gear Junky — Thu, 24 Jul 2014 21:30:49 -0500

So for problem #3 on the In-Class worksheet I worked it through, but didn't have enough time to validate my answer with the rest of my group. I approached it as: $$\int_0^{2\pi} \int_0^{\sqrt{\pi/2}} \int_0^{\cos(r^2)} Z \delta z r \delta r \delta \theta$$ $$=2\pi \int_0^{\sqrt{\pi/2}} \int_0^{\cos(r^2)} Z \delta z r \delta r$$ $$=2\pi \int_0^{\sqrt{\pi/2}} \frac{\cos^2(r^2)}{2} r \delta r$$ And by the Double angle formula I got: $$=\frac{\pi}{2}\int_0^{\sqrt{\pi/2}}r(1+\cos(2r^2)\delta r$$ Using U-sub: $$u=2{r^2}$$ $$\delta u=4r \delta r$$ So I ended up with: $$=\frac{\pi}{8} \int_0^\pi 1+\cos(u) \delta u$$ $$=\frac{\pi}{8}(u+\sin(u)) |_0^\pi$$ Getting an answer of: $$\frac{\pi^2}{8}$$ If anyone got a different answer or can see an error in my math I would greatly appreciate your input.

Graphical explanation

Anonymous — Thu, 24 Jul 2014 10:40:31 -0500

Can someone give a graphical explanation (pictures please) as to how you can find the answer to question 1b of todays worksheet? $$\int_0^1\int_0^{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2-y^2}}1dzdydx$$ No one in my group could really grasp what McClure was saying in class.

Setting up an integral for the volume under a plane

Anonymous — Thu, 24 Jul 2014 10:14:06 -0500

I just want to know if I am doing this correctly. On the worksheet from today, number 6 asks: Set up an integral representing the volume under the plane $x+2y+z=2$ and in the first octant. I came up with the following: $$\int_0^2\int_0^\frac{2-x-z}{2}\int_0^{2-2y-z} dxdydz$$ am I doing this right or am I completely lost?

Spherical and cylindrical problems

Anonymous — Wed, 23 Jul 2014 12:03:06 -0500

I am having a lot of trouble visualizing and understanding how to set up spherical/cylindrical integrals. In particular, I am having trouble with #4 on the homework sheet. "Let $D$ denote the three-dimensional domain above the cone $z=\sqrt{x^2+y^2}$ and inside the sphere $x^2+y^2+z^2<=4$. Evaluate $\int\int\int(x^2+y^2+z^2)dV$." By looking at this, I can guess that spherical coordinates would work well because $x^2+y^2+z^2=P^2$ but past that I am lost. Please help!

In class Problem

Anonymous — Wed, 23 Jul 2014 09:41:37 -0500

In class today my group and I had a little problem solving the second problem written on the board: Evaluate $\int\int\int sin((x^2+y^2+z^2)^{3/2}) dV$ where $D$ is defined as the top half of the solid unit sphere. This is how far we got $$\int_0^\pi\int_0^\pi\int_0^1 (sin((p^2)^{3/2})p^2 sin\phi dP d\phi d\Theta$$ $$\pi\int_0^\pi\int_0^1 sin(p^3)p^2 sin\phi dP d\phi$$ $$u=p^3$$ $$1/3 du=p^2 dP$$ $$\frac{\pi}{3}\int_0^\pi\int_0^1 sin(u) sin\phi du d\phi$$ $$\frac{\pi}{3}\int_0^\pi sin\phi - cos(u) \biggr|_0^1 d\phi$$ $$\frac{\pi}{3}\int_0^\pi 1-cos(1) d\phi$$ $$\frac{\pi}{3} (\phi-\phi cos(1)) \biggr|_0^\pi$$ $$\frac{\pi}{3} (\pi - \pi cos(1) $$ Did we start with the correct domain of integration or did we make a mistake in the calculations?