Find the area of the surface. The part of the sphere
x2 + y2 + z2 = 81
that lies above the plane
z = 4.

Parameterize this surface by

r (u, v) = 9 cos(u) sin(v) i + 9 sin(u) sin(v) j + 9 cos(v) k

with 0 ≤ u ≤ 2π and 0 ≤ v ≤ arccos(4/9)

See the attached sketch to see how we arrive at the upper limit for v.

Take the normal vector to the surface to be

n = ∂r/∂u × ∂r/∂v

n =  -81 cos(u) sin²(v) i -  81 sin(u) sin²(v) j - 81 sin(v) cos(v) k

which has magnitude

||n|| = 81 sin(v)

Then the surface area is

If you're not familiar with surface integrals, you can instead use what's sometimes called the projection method. Let

z = f(x, y) = √(81 - x ² - y ²)

(where we take the positive square root because we're looking at a part of the top half of the sphere)

Projecting the surface down onto the (x, y)-plane, we see that it casts a "shadow" of a disk with radius √(9² - 4²) = √65. (Use the Pythagorean theorem to solve for the missing side of the triangle in the sketch.)

Then the surface S considered above is hovering over the set in the (x, y)-plane,

D = {(x, y) : x ² + y ² ≤ 65}

The area is then

It's easier to compute this integral in polar coordinates, so we take

x = r cos(t )

y = r sin(t )

dx dy = r dr dt

and the region D is given by the set

{(r, t ) : 0 ≤ r ≤ √65 and 0 ≤ t ≤ 2π}

Then the integral would be

Substitute s = 81 - r ² and ds = -2r dr :

is only a line? i think so

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!