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

The time would be 3:00

There are 60 minute per hour. If it is 11:45, it's 25 minutes until 12:10. Once it is 12:10, the next hour (after 50 minutes) would be  1:00. so if you take 11:45 and add the 25 minutes from 3 hours and 25 minutes, it is 12:10 and you are left with 3 hours. 3 hours after 12:10 is 3:10