If the equation of the curve is x2 + y2 = a2 thena)Surface area of the solid generated by revolving about x - axis is 4πa2b)Surface area of the Solid generated by revolving about y - axis is4πa2c)Surface area of the Solid generated by revolving about x - axis is 2πa2d)Surface area of the Solid generated by revolving about y- axis is 2πa2Correct answer is option 'A,B'. Can you explain this answer?

Question

If the equation of the curve is x2 + y2 = a2 thena)Surface area of the solid generated by revolving about x - axis is 4&pi;a2b)Surface area of the Solid generated by revolving about y - axis is4&pi;a2c)Surface area of the Solid generated by revolving about x - axis is 2&pi;a2d)Surface area of the Solid generated by revolving about y- axis is 2&pi;a2Correct answer is option 'A,B'. Can you explain this answer?

Answer

To find the surface area of the solid generated by revolving the curve about the x-axis, we can use the formula for the surface area of revolution:

S = 2π∫[a, b] y * √(1 + (dy/dx)^2) dx

First, let's solve the equation of the curve x^2 + y^2 = a^2 for y:

y^2 = a^2 - x^2

y = ±√(a^2 - x^2)

Since we are only interested in the positive y-values, we can rewrite the equation as:

y = √(a^2 - x^2)

Next, let's find dy/dx:

dy/dx = (-x) / (√(a^2 - x^2))

Now, let's substitute y and dy/dx into the surface area formula:

S = 2π∫[a, -a] √(a^2 - x^2) * √(1 + (-x/√(a^2 - x^2))^2) dx

Simplifying the expression under the square root:

S = 2π∫[a, -a] √(a^2 - x^2) * √(1 + x^2 / (a^2 - x^2)) dx

Simplifying further:

S = 2π∫[a, -a] √(a^2 - x^2) * √((a^2 - x^2 + x^2) / (a^2 - x^2)) dx

S = 2π∫[a, -a] √(a^2 - x^2) * √(a^2 / (a^2 - x^2)) dx

S = 2π∫[a, -a] √(a^2) dx

S = 2π∫[a, -a] a dx

S = 2π * a * [x] [a, -a]

S = 2π * a * (a - (-a))

S = 2π * a * 2a

S = 4πa^2

Therefore, the surface area of the solid generated by revolving the curve x^2 + y^2 = a^2 about the x-axis is 4πa^2.

Answer

Given cuive is

x² + y² = a²

about x-axis

similarly about y- axis

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