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

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.

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.