Recent content by Kerrigoth

This is my first integral. Yes, the integral above represents the volume generated by the area between the line and the parabola. But then I go on to subtract the volume generated by the area of the parabola between y > 0 or if y < -5/2 with the following integral...

My sincerest apologies. I did square it when calculating but mistyped the Tex. Nonetheless, ##\pi\int_{-5/2}^{0}(y^2+\frac{5}{2}y)^2\ dy## does equal ##3.2552\pi##. Being said, my final answer ends up being ##277.545\pi## which is ##18.346\pi## larger than the book's answer.

The pictures provided are normal graphs rotated 90 degrees counter clockwise. The red is the y-axis and the blue is the x-axis. I should have labelled the graph better but can assure they are still correct. I rotated the picture after generating to ease visualization for myself.

Ok, so I'll try that. (again the relevant area whose volume will be counted twice is shaded in pink) So the area I have which includes the double counted volume is which comes from Now I want to subtract the volume (which comes from the area shaded pink) 1. First find the values which...

original graph: I see the problem the pointed out. (I shaded the double counted area pink) The only thing I can think of so far is to shift the graph upwards so there are no negative regions. I thought I'd try: 1. Find the minimum of ##x=y^2+\frac{5}{2}y## ##\frac{\mathrm{d} }{\mathrm{dy}...

Homework Statement The total area between a straight line and the parabola is revolved around the y-axis. What is the volume of revolution? According to the book, the answer is ; My answer comes out to be Homework Equations The Attempt at a Solution 1. Rewrite the second equation in...

The book I'm reading is discussing the physics of semiconductors. I'm having a hard time understanding a passage in section introducing n-type semiconductors. (Phosphorus is used as the impurity) The book says: "At normal operating temperatures, this extra electron breaks its bond with the...

Let's say I place two different size conducting loops inside of a uniform magnetic field. For the sake of the diagram I have the loops share a point, but just treat them as though they are not touching. Then I decrease the value of the magnetic field very quickly. \frac{dB}{dt} =...

The book I am working through is beginning to dive into induced currents and Lenz's Law. When a conductor is moving perpendicular to a magnetic field, the charge carriers inside will experience a magnetic force; This causes the charges to move. Eventually, the electric field from the charge...

Wow, I can't believe I was overlooking that distance factor the whole time...When I was drawing out my situations on paper I wasn't giving the physical distance the right attention because it was just a sketch. You are correct. Thanks!

I'm not disputing the math adds up. What I'm struggling with is identifying what happened after I slid the center plate in between the capacitor. There must have been some transient event that caused the +plate of the capacitor from being able to hold 6+ units of charge, to only 3+ units of...

This is true. But in circuit A the battery is capable of delivering more charge on the +plate of the capacitor attached to it. In circuit A the +plate of the capacitor will hold +6. In circuit B, after you add the center plate, the +plate of the capacitor will only hold +3. What does the center...

Hey daaven, Thanks for welcoming me to PF and thanks for the response! Like you showed in your post, the math very much so works out. But I'm still unclear about what phenomena is going on under the hood that address my original question(s). I'm trying to figure out the phenomena that explains...

First in Circuit A: I hook up a battery with a voltage of V up to a capacitor C1. Once C1 is fully charged it has 6+ on its positive plate and 6- on the negative plate. Given it's fully charged, the +terminal of the battery and the +plate are in electrostatic equilibrium. Since they are both in...