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09/07/2019

Calculus Help? Finding volume by rotating function around x-axis?

QUESTION
Calculus Help? Finding volume by rotating function around x-axis?
I am working on my calculus homework and need help with a few problems involving volume of functions when rotated around the x-axis.

1. The line segment from (0,0) to (6,24) is revolving around the x-axis to form a cone. What is the volume of the cone.

– I have been using the equation y=x/2 which is a line, but I don’t seem to be getting the correct answer. I keep getting 18*pi

2.Find the volume of the solid formed by revolving the region bounded by the graphs of y=x^2+3 and y=12 about the x axis. What is the volume?

– I keep doing this problem the way taught in class but I feel as if I am making a small error that is affecting my answer.

3. Find the volume of the solid formed by revolving the region bounded by the graphs of y=e^x, y=3, and x=0 about the x-axis. What is the volume?

-Having the same problem as the previous one.

4.Let R be the region bounded below by the graph of y=(x+1)/2 and above the graphs of y=5x-4 and y=14-4x. Find the volume of the solid obtained by revolving R about the x-axis.

Thank you in advance for your help! Just helping me figure out how to solve any of the questions would be very helpful!

ANSWER
In these questions, you need good visualisation in 3D to see what intergral(s) you are doing.

(by the way, your first question’s equation is wrong. It needs to be y = x/4, so it will cross (6, 24).)

First, you need to separate the integral into continuous volumes. The only place where this would apply is your last question, where the graphs bounding above meet at x = 2, and where they are bound below by x = 1 and x = 3. So you will need to integrate from 1 to 2 and again from 2 to 3 with the respective equations.

Now comes the integral part. You will need to find the “bigger” integral, do that, and the smaller integral from it. Usually this means integrating the upper-binding graph and subtracting the lower-binding graph.

I’ll do q3 for you as an example:
y = e^x is definitely lower than y = 3. This graph will intercept at x = ln(3). So first we find the cylindrical area from x = 0 to ln(3) (no integral required, just a cylinder), which is area*height
= pi*radius(this is the y value)^2 * height(this is the x value). Imagine the cone lying sideways. Then we have 3^2*pi*ln(3) = 9ln(3)pi.

Next, we follow standard procedures to work out the “hole” in the cylinder created by y = e^x. Now we are integrating from x = 0 to ln(3) of pi*integral(e^(2x))dx. I trust that you can do this part yourself judging from your question.
And that’s it! Subtract what you have here from 9ln(3)pi and you have your answer.

(Edit: I’m more inclined to help people who are aiming for self-improvement rather than just copy-pasting questions without attempting them first – which is why I wrote an essay for you :P)