Question

how do you set up an integral giving the volume of the solid obtained by rotating the region enclosed by:

y=x^2 and y=6x-5 about the x axis.

please help out

Answer 1

1) Decide if you like disks \(\pi\int r^{2}dh\) or washers \(2\pi\int r\cdot h\;dr\).
2) Decide how to cover the entire area of interest.
3) Decide that you will learn more if you do it both ways AND you'll be able to check your won answer.
4) Move on to the next problem feeling awesome at your versatility and understanding.

Answer 2

i dont need to find the integral could you please show me how it looks like i have a test at 6 please

Answer 3

Look really hard at point #1. That is how you set it up.

Answer 4

how would i go about doing a disk?

Answer 5

please could you please set it up i will solve the rest

Answer 6

The only tricky part is finding the points of intersection. Please do that, first. It works nicely to integers. Let's see what you get.

Answer 7

i get x=1 amd x=5

Answer 8

"could you please set it up i will solve the rest"

Solving a couple of integrals is not what you are to be learning. You need to learn how to set them up.

What did you get for those points of intersection?

Answer 9

i used the quadratic formula, i am in 4th year i am just helping someone out i know most of the stuff i just forgot how to do it

Answer 10

Perfect. For future reference, realize that this is x = 1 and x = 5. If, one day, we shoudl decide to turn things around, that correseponds to y = 1 and y = 25. We don;t need that right now, but let's just remember that sometimes we will need to know that.

Okay, now decide which function is uniformly greater than the other on [1,5].

In other words,
is this positive? (x^2) - (6x-5) or
is this positive? (6x-5) - (x^2) on [1,5]

Answer 11

bottom one?

Answer 12

Okay, the rest if formulaic. Just type it as fast as you can think:

\(\pi \int\limits_{1}^{5} (6x-5)^{2}\;dx - \pi \int\limits_{1}^{5} (x^{2})^{2}\;dx\)

or, if you like

\(\pi \int\limits_{1}^{5} (6x-5)^{2} - (x^{2})^{2}\;dx\)

And you need to stare at my iriginal point #1 to see that this is eactly what I said for disks.

Answer 13

It is always amazing to me that such a complicated expression is EXACTLY the same as this one:

\(2\pi\int\limits_{1}^{25}y\cdot\left(\sqrt{y} - \dfrac{y+5}{6}\right)\;dy\)

Seriously, practice doing it both ways. It is a beautiful thing!