(a) A circle has centre at (−2,−1). One point on its circumference is (−2, −3). Find its radius, correct to 1d.p.
(b) If y = x2 for 0 £ x £ 2 is completely rotated round the x – axis, find the volume of the solid of revolution produced.
(a) To find the radius of the circle, we can use the distance formula between the center of the circle and any point on its circumference. The distance formula is:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
In this case, the center of the circle is (-2, -1), and one point on the circumference is (-2, -3). Plugging in these values into the distance formula, we get:
Distance = sqrt((-2 - (-2))^2 + (-3 - (-1))^2)
Simplifying this, we have:
Distance = sqrt((0)^2 + (-2)^2)
Distance = sqrt(4)
Distance = 2
Therefore, the radius of the circle is 2, correct to 1 decimal place.
(b) To find the volume of the solid of revolution produced by rotating the graph of y = x^2 around the x-axis, we can use the method of cylindrical shells. The volume of a cylindrical shell is given by the formula:
Volume = 2π * ∫[a,b] x * f(x) * dx
In this case, the limits of integration are from 0 to 2, as the graph of y = x^2 lies within this range.
So, the volume of the solid of revolution is:
Volume = 2π * ∫[0,2] x * (x^2) * dx
Integrating this expression, we have:
Volume = 2π * [x^4/4] evaluated from 0 to 2
Volume = 2π * [(2^4)/4 - (0^4)/4]
Volume = 2π * (16/4)
Volume = 2π * 4
Volume = 8π
Therefore, the volume of the solid of revolution produced is 8π units cubed.