A community for students.
Here's the question you clicked on:
 0 viewing
CoconutJJ
 one year ago
The bounded area by the graph y=sqrt(x), y=x4 and the x axis is rotated along the x axis to form a solid. Use Integration By Shells to find the Volume of the Solid.
CoconutJJ
 one year ago
The bounded area by the graph y=sqrt(x), y=x4 and the x axis is rotated along the x axis to form a solid. Use Integration By Shells to find the Volume of the Solid.

This Question is Closed

misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.0This might help utilize the "Shell" method; https://www.khanacademy.org/math/integralcalculus/solid_revolution_topic/solid_of_revolution/v/solidofrevolutionpart5

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435377786774:dw This is the bounded area

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0Wait so how does the integral change when you rotate it around the x axis instead of y?

misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.0https://www.wyzant.com/resources/lessons/math/calculus/integration/finding_volume

misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.0With "washer method"

misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.0To rotate about the yaxis, swap the roles of x and y throughout the above discussion. To rotate about a different axis, change coordinates so that the axis of rotation is one of the coordinate axes, and then do the computation.

misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.0Follow rules of "shell" and "washer" method to guide throughout finding volume

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0been a while, and I was never great at these so take this with a grain of salt. I believe you have to change the formulas around and integrate in terms of y. dw:1435378239112:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and then the limits will be the y coordinates of the intersection points

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0so then the thickness of the cylinder is dy?

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435378522268:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0@peachpi it's been a while for me too.

misssunshinexxoxo
 one year ago
Best ResponseYou've already chosen the best response.0Peachpi seems absolutely right; this is a tough area

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is a solid. so there's no hole, right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the area lies on the axis it's being rotated on

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0but eventually we have to integrate this thing... who is going to be a and who is going to be b on the integral symbol ?

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0well they intersect when x=6.562

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, but the integral is in y. a is 0. b you solve y² = y + 4 to get

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0How is the integral with respect to y ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0maybe it has something to do with the x axis being rotated so we can't use that? I have no idea x.x

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0oh I see where the a and b for the integral is dw:1435378968903:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0those intersection points are the values for a and b

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435378996664:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0well a = 0 is correct.. but for the b we need to find the value of that intersection point but as @peachpi mentioned we have to solve " y² = y + 4 to get"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think it is simpler to do it horizontally than vertically, which would require two integrals.

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0hmmm y^2y4 = 0 and then factor ? OH NO! I DO NOT WANT TO DO 2 INTEGRALS

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0(1)^24(1)(4) = 14(4) = 1+ 16 = 17 and it's not a perfect square.. dahhhh

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What is the point of intersection?

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435379165728:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0you graph the two equations first. and then they should have common points.

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435379200509:dw

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0with 0 being one of them

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0they're not nice numbers. about (6.56, 2.56)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435379227291:dw

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0So if we use the shell method from 0 to 4, we are basically using the disc method?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0double integrals.. double the anti derivatives :S

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1435379315446:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ \int dV = 2\pi\int rh~dy = 2\pi\int \limits_0^{y_1} y(f(y)g(y))~dy \]

CoconutJJ
 one year ago
Best ResponseYou've already chosen the best response.0THANK YOU! I think I can do it from here now :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The volume of a cylinder is: \[ V = \pi r^2h \]When you differentiate, it is: \[ \frac{dV}{dr} = 2\pi r h \implies \int dV = 2\pi\int rh~dr \]That's how I think about it.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.