find the volume of the solid formed by revolving the region bounded by y=x^2, y=0, and x=2 about the y axis

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

- katieb

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- anonymous

Do you know about multivariable calculus perhaps? Or do you want the easy-not-understanding-what-you're-doing way

- anonymous

The answer is 8pi

- anonymous

No, I am supposed to use the Disk Washer method.
I can do it with antiderive(2-y) to get the right answer but I'm pretty sure that's not following th formula I'm given which is outer radius minus inner radius.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

Disk Washer?

- anonymous

There's a hole in the solid.. so it's called Washer Method.

- anonymous

So how does this method work?

- anonymous

I do antiderivative from a-b in terms of x or y (y in this case because I rotate about y axis)
pi (outer radius)^2 - (inner radius)^2

- anonymous

\[\int\limits_{a}^{b} \pi (outer radius)^2 -(innerradius)^2 dx\]

- anonymous

I don't how that works, I'm afraid.

- zepdrix

|dw:1359838432884:dw|

- zepdrix

|dw:1359838561883:dw|So let's look at one slice.

- zepdrix

Oh you're doing the washer/disk method, my mistake..
Lemme slice that differently.

- zepdrix

|dw:1359838731527:dw|Ok this type of slice will give us a Disk.

- zepdrix

|dw:1359838764034:dw|So we want to get the Volume of this disk.

- zepdrix

The outer radius appears to be the line x=2.
And the inner radius our function which is in terms of x, we'll need it in terms of y to integrate since we sliced in the y direction (dy thickness)

- zepdrix

\[\large y=x^2 \qquad \rightarrow \qquad x=\sqrt y\]

- zepdrix

\[\large V=\pi\left[(2)^2-(\sqrt y)^2\right]dy\]

- zepdrix

Then to find the total volume in this enclosed area, we Integrate (add up all the slices) from one intersecting point to another.
Where do they intersect? Ummm looks like.. y=0 and y=4?

- zepdrix

Simplifying things down gives us,
\[\large \pi \int\limits_0^4 4-y \;dy\]

- zepdrix

Imma check my work real quick to make sure I didn't make a mistake somewhere. A little tired, it's possible. lol

- zepdrix

Yah I think that's right.
Confused about any of that?
I went through it a little sloppy D:

- anonymous

It's right. I realize what I did wrong! I had the wrong intersections/points.
I did instead from -2 to 2.. not realizing that it was based on terms of y .
thank you so much!

- zepdrix

Oh cool c:

Looking for something else?

Not the answer you are looking for? Search for more explanations.