Area and Volumes of Solid of Revolution

[Oman]

1. find the area common to r=1+cos@ and r=3^(1/2) sin@

2. find the volume generated by rotating the region bounded by
(x-1)^2 + (y-2)^2 = 4 around
a. x axis
b. y -axis
c. x = 3
d. y = 4

 

[AiRAVATA]

So? What have you done?

 

[tacman]

1. To find the area common to r=1+cos@ and r=3^(1/2) sin@, we first need to set up the integral. Since we are dealing with polar coordinates, the area formula is given by A=1/2∫(r)^2 d@. We can rewrite both equations in terms of @ by substituting x=rcos@ and y=rsin@. This gives us r=1+cos@ and r=√3sin@.

To find the common area, we need to find the points of intersection between the two curves. Setting them equal to each other, we get 1+cos@=√3sin@. Solving for @, we get @=π/3 and @=5π/3.

Now, we can set up the integral as A=1/2∫(√3sin@)^2 d@ from @=π/3 to @=5π/3. Evaluating the integral, we get A=π/2-√3/4 square units.

2. To find the volume generated by rotating the region bounded by (x-1)^2 + (y-2)^2 = 4, we need to use the formula V=∫π(r)^2 dy (if rotating around the x-axis) or V=∫π(r)^2 dx (if rotating around the y-axis).

a. For rotating around the x-axis, we can rewrite the equation in terms of x by substituting x=r+1 and y=r+2. This gives us r=√(3-x^2). Plugging this into the formula, we get V=∫π(√(3-x^2))^2 dy from y=0 to y=3. Evaluating the integral, we get V=9π/2 cubic units.

b. For rotating around the y-axis, we can rewrite the equation in terms of y by substituting x=r+1 and y=r+2. This gives us r=√(3-y^2). Plugging this into the formula, we get V=∫π(√(3-y^2))^2 dx from x=0 to x=3. Evaluating the integral, we get V=9π/2 cubic units.

c. For rotating around the line x=3,