The volume of the solid generated by the revolution of r = 2a cos θ about the initial line is given by:a)2/3πa3b)4/3πa3c)8/3πa3d)None of theseCorrect answer is option 'B'. Can you explain this answer?

Question

The volume of the solid generated by the revolution of r = 2a cos &theta; about the initial line is given by:a)2/3&pi;a3b)4/3&pi;a3c)8/3&pi;a3d)None of theseCorrect answer is option 'B'. Can you explain this answer?

Answer

To find the volume of the solid generated by the revolution of r = 2a cos(theta), we can use the method of cylindrical shells.

The equation r = 2a cos(theta) describes a cardioid shape. To find the limits of integration for theta, we need to determine where the cardioid intersects the x-axis. Setting r = 0, we have 2a cos(theta) = 0, which gives cos(theta) = 0. This occurs when theta = pi/2 and 3pi/2.

Now, let's consider a small segment of the cardioid between two values of theta, denoted as theta1 and theta2. The radius of this segment is r = 2a cos(theta1) = 2a cos(theta2). The height of the cylindrical shell is given by the difference in the values of theta, which is delta(theta) = theta2 - theta1. The circumference of the cylindrical shell is given by 2πr = 4πa cos(theta).

The volume of each cylindrical shell is then given by V = 2πa cos(theta) (theta2 - theta1). To find the total volume, we need to integrate this expression over the range of theta from pi/2 to 3pi/2:

V = ∫[pi/2, 3pi/2] 2πa cos(theta) d(theta).

Using the identities for the integral of cos(theta), we have:

V = 2πa [sin(theta)]_[pi/2]^[3pi/2] = 2πa (sin(3pi/2) - sin(pi/2)).

Since sin(3pi/2) = -1 and sin(pi/2) = 1, we have:

V = 2πa (-1 - 1) = -4πa.

Therefore, the volume of the solid generated by the revolution of r = 2a cos(theta) is -4πa. Note that the negative sign indicates that the solid is oriented below the x-axis.

Answer

4πa²

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