as level p1 dy/dx 👍👍👍?

y=sqr(8x-x^2)

(i) dy/dx=(8-2x)/[2sqr(8x-x^2)]=(4-x)/sqr(8x-x^2)

is the required expression.

dy/dx=0=>x=4 & y(4)=sqr[8(4)-4^2]=4=>the required stationary point=

(4,4).

(ii) y=0=>sqr(8x-x^2)=0=>x(8-x)=0=>x=0 or x=8

Thus, the required volume is

........8

V=piSy^2dx=>

........0

........8

V=piS(8x-x^2)dx=>

........0

........8

V=pi[4x^2-(x^3)/3]=>

........0

V=256pi/3~268.1.

y = √(8x − x²)

x² + y² - 8x = 0, y ≥ 0

The curve is the semicircle having center (4, 0), radius 4, and bounded below by the x-axis.

The slope of any radius to point (x, y) on the curve is y/(x - 4).

The tangent through that point has slope (4 - x)/y, being perpendicular to the radius.

dy/dx = (4 - x)/y = (4 - x)/√(8x − x²)

The stationary point must be (4, 4), the point on the curve directly above the center.

The axis of rotation is the diameter of the semicircle, so the solid of revolution is simply a sphere of radius 4.

volume = (4/3)π(4)³ = 256π/3

I agree with the responder identified as "?", that the stationary point of the curve occurs where x = 4, and that the coordinates of that point are (4,4).

For the volume I disagree with Vaman's assertion that the volume is the integral of 2*pi*y*dx, it is not.  It is the integral of pi*y^2*dx from x = 0 to 8, so it is the integral of

pi(8x - x^2) dx.

After integration you have pi[4x^2 - (1/3)x^3], 

and when you plug in the limits you get

pi[256 - (1/3)512] 

= pi[(256*3 - 256*2)/3] = 256*pi/3, around 290.

The answer is as follows:

y = √(8x-x²) ⇒

y = (8x-x²)¹′²

(i) dy/dx = ½ (8x-x²)⁻¹′² (8-2x)

....// stationary points occur when dy/dx = 0

....½ (8x-x²)⁻¹′² (8-2x) =

........(8-2x)

....-------------- = 0 ⇔

....2 (8x-x²)¹′²

....8-2x = 0 ⇔

....2x = 8

....x = 4

....// Evaluate y = √(8x-x²) at x=4

....// to find coordinates of stationary point .

....// When x = 4, y = ±4, however we can

....// ignore y= -4 because of the radical

....// sign: y = √(8x-x²)  > 0. So,

....stationary point = (4,+4)................ANS

-------------

(ii) Use integration to find the volume

y=z^(1/2)

dy/dz= 1/2 z^(-1/2), Use y= z^n, dy/dx= n z^(n-1) dz/dx

z=(8x-x^2), dz/dx= 8-2x

dy/dx= 1/2 (8x-x^2)^(-1/2) (8-2x)

Ans 2.

y=0 at x=0 x=8.

The volume will be = int 2 pi y dx with limits x=0 and x=8

int y dx= int (8x-x^2)^1/2 dx=

I give hint  to integrate. put x= 8 sin^2 z. dx= 16 cos z sin z dz. And integrate it. Put the limits. You will get the answer.