integral in terms of k?

2008-11-16 10:51 pm

Let R be the region of the first quadrant bounded by the x-axis and the curve y = kx - x^2, where k > 0
(A) in terms of k, find the volume produced when R is revolved around the x-axis
(B) in terms of k, find the volume produced when R is revolved around the y-axis
(C) find the value of k for which the volumes found in (A) and (B) are equal

i got the integrals...but i need to know if i'm on the right track or how close I am to the right answer :) thanks

回答 (1)

[匿名]

2008-11-17 10:24 pm

✔ 最佳答案

(A) This is relatively easy to do by the method of disks, where a typical disk has radius x(k-x) and width dx. Thus the volume of the solid of revolution is given by
V = pi int_{x=0}^{x=k} x^2(k-x)^2 dx
= pi int_{x=0}^{x=k} x^2(k^2 - 2kx + x^2) dx
= pi [k^2(x^3/3) - 2k(x^4/4) + x^5/5]_{x=0}^{x=k}
= pi [k^5/3 - k^5/2 + k^5/5]
= pi(k^5/30)[10 - 15 + 6]
= pi(k^5/30).

(B) Again, you could probably do this by the method of disks, but the radii of the outer and inner disks in this case are rather awkward functions involving the (two) solutions of x(k-x) = y (in terms of x). So let's use the method of shells instead, where the typical shell has radius x and height x(k-x). Then the volume of the solid of revolution is given by
V = 2pi int_{x=0}^{x=k} x^2(k-x) dx
= 2pi [kx^3/3 - x^4/4]_{x=0}^{x=k}
= 2pi [k^4/3 - k^4/4]
= pi(k^4)/6.

(C) Will let you do this part.

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