Question

Asked Feb 8, 2019

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26. Let R be the region bounded by(y = x^2) and y = 4. Compute the volume of the solid formed by revolving R about the given line.

y = 4 (b) the y-axis (c) y = 6

(d) y = –2 (e) x = 2 (f) x = –4

(b) this is what figured out and got from symbolab

pi\int _{-2}^2\:\left(4\right)-\left(x^2\right) or 33.51032

Step 1

In order to understand the given question, please see the picture on the white board.

The parabola represents the curve y = x^{2}. AB is the line y = 4 that cuts the parabola at point A and B.

x coordinate of the points A and B are -2 and 2 respectively.

The shaded portion is the area enclosed between the line y=4 and the curve y = x^{2}. Now consider a thin strip of area shown as double hashed. If the entire graph is rotated across y=4 i.e. line AB, the thin stip of area will give rise to a circular area on full rotation.

The radius of this circle will be, R = 4 - x^{2}. Please see what R is pictorially.

Step 2

Area of this typical circle, A = pi.(R^{2})

If we move this circle by an amount dx across the x axis, we will end up getting a small volume, dV = A.dx

In order to get the entire volume we will have to integrate dV = A.dx over the entire range of x i.e. from point B to A.

Step 3

Please see the evaluation of volume on white board.

Since the integran is an even ...

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