Take the region under the graph y=cos(x) and above the graph y=sin(x) for 0 1)Fi

Question

Take the region under the graph y=cos(x) and above the graph y=sin(x) for 0

1)Find the volume of the solid of revolution that is generated when this region is rotated about the x-axis.

2)Find the volume of the solid of revolution that is generated when this region is rotated about the line y=2.

3)Set up the integral you will need to evaluate to find the volume of the solid revolution that is generated when this region is rotated about the y-axis.

4)Set up the integral you will need to evaluate to find the volume of the solid revolution that is generated when this region is rotated about the line x=-3.

Explanation / Answer

1) The outer radius is: cos(x) - 0 The inner radius is: sin(x) - 0 So, the area of each slice is: pi*(cos^2(x) - sin^2(x)) And the integral is: integrate( pi*(cos^2(x) - sin^2(x)), dx, 0, pi/4) This integral evaluates to : pi/2 B) The outer radius is: 2 - sin(x) The inner radius is: 2-cos(x) The area of each slice is: pi*((2-sin(x))^2 - (2-cos(x))^2) So, the integral is: integrate ( pi*((2-sin(x))^2 - (2-cos(x))^2) , dx, 0, pi/4) This integral evaluates to: (8sqrt(2) - 9)*pi/2 C) We would do this using the shells method. The radius for each shell is: x - 0 The height of each shell is: cos(x) - sin(x) The area of each shell is: 2pi*r*h 2pi(x)(cos(x) - sin(x)) So the integral would be: integrate( 2pi(x)(cos(x) - sin(x)), dx, 0, pi/4) 4) This is the same as the last one except the the radius is: x- (-3) = x + 3 So the integral is: integrate( 2pi(x + 3)(cos(x) - sin(x)), dx, 0, pi/4)

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