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# Find the volume of the solid obtained by rotating the region bounded by y = 2 and y = 1 + (x-2)2 about the line x = -1

Calculus question

Q: Find the volume of the solid obtained by rotating the region bounded by y = 2 and y = 1 + (x-2)2 about the line x = -1

1)Find the two points where y = 2 line intersects with the parabola as well as the vertex point of the parabola. Setting y to 2 and solving for x in the equation yields x = [1,3]. The vertex form of the parabola equation is:
y = a(x - h)^2 + k. The vertex point is (h,k) or (2,1).

These three points will help us find the limits of integration.

2) The total volume can be divided into a series of partial circles or discs stacked on top of each other forming the whole volume. The area of each circle is pi*m^2-pi*n^2 where m and n are the two radii bounding each disc. These radii are the distances from the line x = -1 to the two sides of the parabola.

3)To find the total volume the areas of the discs stacked on top of each other are summed together. To do this integrate the areas from the bottom to top. Our limits of integration will be from the parabola vertex point to the line y = 2. Therefore the limits of integration will be from 1 to 2.

4)Consequently the formula for finding the total volume is V = ∫(pi*m^2 - pi*n^2)dy from 1 to 2.

5)The next step is to find the formula for the two radii. Since we are measuring distances from the line x = -1. The parabola equation must be solved in terms of y. This will give us:
x=+-sqrt(y-1) + 2. Subtracting this from x = -1 yields: +-sqrt(y-1) + 3. Where the positive root is the outer radius and the negative root is the inner radius.

6)Therefore the final equation is: V = ∫((pi*(sqrt(y-1) + 3) ^2 - pi*(-sqrt(y-1) +3)^2))dy from 1 to 2.

On solving the above integration---
8pi [ (y-1)^3/2 ].     limits of integration as 1 to 2

=8pi

Total enclosed area= 8pi

Surendra
Find the volume of the solid obtained by rotating the region bounded by y = 2 and y = 1 + (x-2)^2 about the line x = -1.

Find intersections of y = 2 and y = 1 + (x-2)^2:

2 = 1 + (x-2)^2

(x-2)^2 = 1

x-2 = ± 1

x = 2 ± 1 = 1, 3

Every vertical line segment between x = 1 and x = 3 with length 2 – (1 + (x-2)^2) represents the height of a cylindrical shell with axis x = –1, surface area = hC, and volume = hC dx. Just add ‘em up.