Find the area of the surface generated by these equations arounf the y axis

[MozAngeles]

Homework Statement

Find the area of the surface generated by these equations around the y axis
x=t²+1/(2t) y=4√t ; 1/√2≤t≤1

Homework Equations

S=∫2$$\pi$$ x√(dx/dt)²+(dy/dt)²)dt

The Attempt at a Solution

dx/dt=2t-1/2t² squaring that, 4t^2-2/t+1/(4t4)

dy/dt=2/√t squaring that , 4/t

plugging this intot he formula
S= 2$$\pi$$∫ (t²+1/(2t)√(4t²-2/t+1/(4t⁴)+4/t)
S= 2$$\pi$$∫ (t²+1/(2t) √((2t+1/(2t²))²
S= 2$$\pi$$∫ (t²+1/(2t) (2t+1/(2t²))

so this is where I am stuck u sub almost works here, if it weren't for that pesky negative sign. So could someone help me maybe they noticed a mistake in my work. anything thank you much.

 

[Mark44]

I'm having some difficulty trying to decipher some of your work. To help other readers and myself, I am fixing your LaTeX.

Tip: If you use tex tags, use one pair for an entire equation.

Homework Statement

Find the area of the surface generated by these equations around the y axis
x=t²+1/(2t) y=4√t ; 1/√2≤t≤1

Homework Equations

S=∫2$$\pi$$ x√(dx/dt)²+(dy/dt)²)dt

Is this your integral?
$$S = 2\pi \int_a^{b} x\sqrt{(dx/dt)^2+(dy/dt)^2}dt$$

The Attempt at a Solution

Fixed the equation below.

dx/dt=2t-1/2t² squaring that, 4t²-2/t+1/(4t⁴)

dy/dt=2/√t squaring that , 4/t

plugging this intot he formula
S= 2$$\pi$$∫ (t²+1/(2t)√(4t²-2/t+1/(4t⁴)+4/t)
S= 2$$\pi$$∫ (t²+1/(2t) √((2t+1/(2t²))²
S= 2$$\pi$$∫ (t²+1/(2t) (2t+1/(2t²))

$$S = 2\pi \int_{1/\sqrt{2}}^1 (t^2+1/(2t) (2t+1/(2t^2))dt$$

so this is where I am stuck u sub almost works here, if it weren't for that pesky negative sign. So could someone help me maybe they noticed a mistake in my work. anything thank you much.

 

[Mark44]

You don't need a substitution. Just multiply the two factors and then integrate. For the integrand, I get 2t³ + 5/4 + 1/(8t³)