Find the area of one side of the "winding wall" standing orthogonally on the curve y= x^2 , 3<= x <=5 and beneath the curve on the surface f(x, y) = x + √y.

The required area is 131.6626

Step-by-step explanation:

Given that;

equation of the curve;  y=x², 3≤ x ≤5

and equation of the surface; f(x,y) = x + √y

Find the Area A = ∫_c ( x + √y) dS

Find the parametrization of the curve.

so let x = a then y = a²

then dx/da = 1, dy/da = 2a

since 3≤ x ≤5 and x = a hence; 3≤ a ≤5

so the area of one side of winding wall will be;

A = ∫_c ( x + √y) dS

= ⁵∫₃ ( a + √a²) √[(dx/da)² + (dy/da)²] da

= ⁵∫₃ ( a + a ) √[(1)² + (2a)²] da

= ⁵∫₃2a√(1 + 4a²) da

= [1/6(4a² + 1 )^3/2)]₃⁵

= - 37/6×√37 + 100/6×√101

= -37.5103 + 169.1729

= 131.6626

Therefore the required area is 131.6626

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)