Hey there! Sign in to join this conversationNew here? Join for free
x Turn on thread page Beta

P3 another parametric question watch

Announcements
    • Thread Starter
    Offline

    0
    ReputationRep:
    The curve C is given by the equations x=2t y=t^2

    (a) Find an equation of the normal to C at the ppoint on C where t=3

    well answer to that is y=-1/3x+11 i got that

    The normal meets the y-axis at point B. The finite region R is bounded by part of the curve C between the origin O and P, and the line OB and OP

    (b) Show the Region R, together with its boundaries, in a sketch (below is what i think it loks like and i think it is right)

    The region R is rotated through 2pi about y-axis to form solid S.

    (c) Using integration, and explaining each step in your method, find the volume of S, giving your answer in terms of pi.
    i cant get the right answer for that it is 186pi.

    ONe general question when it says that the area is rotated through 2pi would you multiply the integral by 2pi or just pi?
    Attached Images
     
    Offline

    1
    ReputationRep:
    (Original post by lexazver203)
    (c) Using integration, and explaining each step in your method, find the volume of S, giving your answer in terms of pi.
    i cant get the right answer for that it is 186pi.

    ONe general question when it says that the area is rotated through 2pi would you multiply the integral by 2pi or just pi?
    Can't really make out the sketch, for c) is it rotated around the x or y axis.

    (Its only one pi).
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Bhaal85)
    Can't really make out the sketch, for c) is it rotated around the x or y axis.

    (Its only one pi).

    y-axis
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by IntegralAnomaly)
    here's the volume:
    what does that come to? i dont really understand how o work it out.
    Offline

    1
    ReputationRep:
    (Original post by lexazver203)
    The curve C is given by the equations x=2t y=t^2

    (a) Find an equation of the normal to C at the ppoint on C where t=3

    well answer to that is y=-1/3x+11 i got that

    The normal meets the y-axis at point B. The finite region R is bounded by part of the curve C between the origin O and P, and the line OB and OP

    (b) Show the Region R, together with its boundaries, in a sketch (below is what i think it loks like and i think it is right)

    The region R is rotated through 2pi about y-axis to form solid S.

    (c) Using integration, and explaining each step in your method, find the volume of S, giving your answer in terms of pi.
    i cant get the right answer for that it is 186pi.

    ONe general question when it says that the area is rotated through 2pi would you multiply the integral by 2pi or just pi?
    For b) I just drew Y=(x^2)/4 and normal tangent.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Bhaal85)
    For b) I just drew Y=(x^2)/4 and normal tangent.

    but it is also intersected by the normal which crosses it at point P
    Offline

    1
    ReputationRep:
    (Original post by lexazver203)
    but it is also intersected by the normal which crosses it at point P
    That is what I just said. But yes, the graph looks correct.
    Offline

    2
    ReputationRep:
    (Original post by lexazver203)

    (c) Using integration, and explaining each step in your method, find the volume of S, giving your answer in terms of pi.
    i cant get the right answer for that it is 186pi.

    ONe general question when it says that the area is rotated through 2pi would you multiply the integral by 2pi or just pi?
    volume = pi [int (x^2) dy]

    if you change into cartesian (you could also keep it in 't', but would have to change the limits), y= (x^2)/4, so
    x^2 = 4y

    V = pi[int 4y dy] between the limts 0 and 9
    V = pi(2y^2)
    sub limits, to get volume of curve rotated about y-axis as 162pi

    for the other shape, you can see that if it is rotated, it will become a cone, where: r=6, and h=2 (or you could also integrate the line between the limits 9 and 11).

    so v2 = (pi/3)(6^2)(2) = 24pi

    so total volume is 162pi + 24pi = 186pi units cubed
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Bhaal85)
    That is what I just said. But yes, the graph looks correct.
    when u integrate would you do it in 2 parts? i mean do triangle and area enclosed by the curve? i did it that way but still cant get the right answer.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by lexazver203)
    when u integrate would you do it in 2 parts? i mean do triangle and area enclosed by the curve? i did it that way but still cant get the right answer.


    what is the formula for the volume of a cone
    Offline

    1
    ReputationRep:
    (Original post by lexazver203)
    when u integrate would you do it in 2 parts? i mean do triangle and area enclosed by the curve? i did it that way but still cant get the right answer.
    Do them in two seperate things. Get the equations into a cartesian form i.e x^2/4=y therefore X^2=4y and integrate between 9 and 0 to get 162pi and then add the area of the triangle.
    Offline

    2
    ReputationRep:
    (Original post by lexazver203)
    what is the formula for the volume of a cone
    (1/3)*pi*(r^2)(h)
 
 
 
Poll
Do you like carrot cake?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.