A function is given as f(x)=x∙√(x^2+1) ; x≥0?

Question

A:Find the Area bounded by the function, the x-axis and x=3
B:Find the Volume when the area is rotated around the x-axis 
C: Find the volume of the area rotated around the y-axis (fnint command on TI84 allowed) 
D: Find the volume when the area between the function the y-axis and y=3∙√10, is rotated 70 degrees around the y-axi

cidyah · Answer

A)
∫  x sqrt(x^2+1) dx

Let u= x^2+1
du = 2x dx
x dx = (1/2) du

∫  x sqrt(x^2+1) dx = (1/2) ∫  sqrt(u) du
= (1/2) ∫  u^(1/2) du
= (1/2) u^(1/2 +1)/(1/2 +1)
= (1/2)(2/3) u^(3/2)
= (1/3) u^(3/2)

replace u by sqrt(x^2+1)
= (1/3) (x^2+1)^(3/2)

F(x) = (1/3) (x^2+1)^(3/2)
F(3) = (1/3) (3^2+1)^(3/2)
F(3) = (1/3) (10)^(3/2)
F(0) = (1/3)

Area = (1/3) (10^(3/2) - 1) = 10.2076

B)
Volume = pi ∫  y^2 dx
= pi ∫  x^2 (x^2+1) dx , 0< x < 3
∫  x^2 (x^2+1) dx =  ∫  x^4 dx +  ∫  x^2 dx

= (1/5) x^5 + (1/3) x^3
F(x) = (1/5) x^5 + (1/3) x^3
F(3) = 288/5
F(0) = 0

Volume = 288pi/5

C)
Use the shell method:
radius = x
height = x sqrt(x^2+1)

Volume = 2pi ∫  x x sqrt(x^2+1) dx
= 2pi ∫  x^2 sqrt(x^2+1) dx , 0 < x < 3
https://gyazo.com/497536d9b8c15901cd5916663d9dc7c9

Use the T184 calculator. It's easier.

? · Answer

A.  Integral from x = 0 to 3 of
[x*sqrt(x^2+1)] dx.
Let u = x^2 + 1, so du = 2x dx, and your integrand becomes
(1/2)u^(1/2) du.
After integration you have (1/3)u^(3/2) = (1/3)(x^2 + 1)^(3/2), so the result is
(1/3)[10^(3/2) - 1] = about 10.2.

B.  Do this by disks.  The area of a disk is pi*y^2 = pi*x^2*(x^2 + 1) = pi(x^4 + x^2).
Integral from x = 0 to x = 3 (??? -- in "B" they don't mention the right-hand boundary!)
of pi(x^4 + x^2) dx = pi[(1/5)x^5 + (1/3)x^3] = pi(243/5 - 9) = pi*198/5.

C.  I haven't got a TI-84, so you'll have to cope with this one.  By shells, you'd have
the integral from x = 0 to 3 of 2*pi*x*[x*sqrt(x^2+1)] dx or 2*pi*x^2*sqrt(x^2+1) dx.
Trying to do it by letting x = tan(u) is possible, but VERY tedious, so I recommend the calculator method suggested.

D.  If the "70 degrees" weren't in this problem, the full volume that you get by rotating the upper triangle around the 
would be found by considering the volume of the cylinder r = 3, h = 3*sqrt(10), and then subtracting the answer from part C.  Once you have that difference, you could cope with the 70 degrees -- just multiply the volume by 7/36.

A function is given as f(x)=x∙√(x^2+1) ; x≥0?

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