That's nice, but u haven't solved question 2 and the answer to question 1i is 3/2 not 3/3

Mathematics

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Answers

2 parts
a

24πa324\pi a^3

b

48πa248\pi a^2

Step-by-step solution

4 steps

My apologies if there was any confusion.

For Question 1(i), the calculated length of the curve C was indeed 32\frac{3}{2}.

L=[32x2/3]01=32(12/3−02/3)=32L = \left[\frac{3}{2} x^{2/3}\right]_{0}^{1} = \frac{3}{2} (1^{2/3} - 0^{2/3}) = \frac{3}{2}

The final answer for Question 1(i) is 32\boxed{\frac{3}{2}}.

I did provide solutions for Question 2, parts a) and b). Here they are again for clarity:

Part 2

A square of side 2a2a is rotated through 2π2\pi about a line in its plane which is distant 3a3a from the centre O of the square. Use theorems of Pappus to find:

Part a

24πa324\pi a^3

The Volume of the Solid swept out by the Square.

  1. Identify the area of the square.

    Side length s=2as = 2a.
    Area A=s2=(2a)2=4a2A = s^2 = (2a)^2 = 4a^2.
    The centroid of the square is its geometric center.

  2. Apply Pappus's First Theorem.

    The distance of the centroid from the axis of rotation is d=3ad = 3a.
    The volume VV is given by V=2πdAV = 2\pi d A.

    V=2π(3a)(4a2)V = 2\pi (3a) (4a^2) V=24πa3V = 24\pi a^3

    The volume of the solid swept out by the square is 24πa3\boxed{24\pi a^3}.

Part b

48πa248\pi a^2

The area of the Surface of the Solid.

  1. Identify the perimeter of the square.

    Perimeter P=4s=4(2a)=8aP = 4s = 4(2a) = 8a.
    The centroid of the perimeter is its geometric center.

  2. Apply Pappus's Second Theorem.

    The distance of the centroid from the axis of rotation is d=3ad = 3a.
    The surface area SS is given by S=2πdPS = 2\pi d P.

    S=2π(3a)(8a)S = 2\pi (3a) (8a) S=48πa2S = 48\pi a^2

    The area of the surface of the solid is 48πa2\boxed{48\pi a^2}.

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