Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis for y=x y=0 and x=2

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.

A community for students.

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis for y=x y=0 and x=2

Calculus1
See more answers at brainly.com
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 this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

First, draw the graphs and identify the enclosed area. Have you done that already?
Yes
|dw:1434058993462:dw|

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Since we are revolving the shell around the y-axis, we will be using the example i circled. |dw:1434059042077:dw|
Makes sense?
yeah! except your axis kind of confuse me I thought the y and x were switched
But yeah I understand.
But before anything else. You will analyze the rectangle and see what function along the x-axis is higher and what functions where the x-axis is lower. |dw:1434059232453:dw|
ok I think I get it. The one thing im really confused on is the integral. do you use the numbers on the x-axis for the integral?
\[\int\limits_{0}^{3} 2\pi (x ^{2})\] so like this ?
i used that and then plugged the integral in so i got \[18\pi \] as my answer is that right. I think we may have learned it a different way because we never used that Big X little x thing we use a set equation that combines circumference and height
|dw:1434059661157:dw|
I forgot the mention, the intervals are based on where the spot is enclosed on the y-axis because we have dy and everything should be based on Y.
First step: Let's identify the intervals. What do you think the intervals are?
i gave you the wrong one by accident the problem says y=0 to x=3 so i think those are the intervals
i told you 2 but its 3 sorry about that
okay
So the intervals were right. But what did you get for the big x and the small x?
I don't know because the formula we learned dose not have that anywhere in there so i don't understand that.
I honestly thought i had it right and just wanted to double check my answers and make sure but now im super confused
\(\large a(x)=2\pi(radius)(height)\)
Yeah!
|dw:1434060538490:dw| \(\large a(x)=2pi(y)(X-x)\)
\[\Large \int\limits_{0}^{3}2 \pi y(X-x)dy\]
ok i still don't get the X-x thing i thought it was just the equation which is in this case just x but then u times that by x from 2pi move the 2 pi out front because its a constant take the anti derivative of the x^2 so then you end up with \[2\pi \int\limits_{0}^{3} 1/3 x^{3}\]
|dw:1434060977099:dw| The radius is y, and the height is (X-x)
thanks for helping btw i know its a pain
|dw:1434061121877:dw| The big X is x=3 and the small x is y=x
Makes sense? Hope I didn't confused you...
ok yeah that makes sense.
ok! so then the next step is the anti derivative of that?
Yes.
Use the fundamental theorem of calculus to find the definite integral. \[\Large \int\limits_{a}^{b}f(y)dy=F(b)-F(a)\]
so 9pi or am i still not getting it?
omg! ok thank you so much!!
Np!
This calculator is designed to make the calculation of volumes easier without remembering the difficult formulas of volume. Try http://www.acalculator.com/volume-conversion-calculator.html

Not the answer you are looking for?

Search for more explanations.

Ask your own question