Part 1- Find the area of the region enclosed by the curve with equation y=tanx , the x-axis and the lines x= 0 and x= 1/3pie
Part 2- Find the volume generated when this area is rotated about the x-axis.
Part 2- Find the volume generated when this area is rotated about the x-axis.
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Part 1: Easy. Just integrate tan(x) from x=0 to x=1/3 pi. That represents the area you specified.
Part 2: After this rotation is made, you will have an infinite number of increasing radius cylinders on the same interval in part 1 (0 to 1/3 pi). So what is the volume of a cylinder? (pi)(r)^2 (height of cylinder). We know pi, and the radius is represented by the function tan(x). And the height of the cylinder (which in theory is infinitely small) is represented by the dx in our integral. So:
A definite integral from 0 to 1/3 pi of (pi (tanx)^2 dx. Solve the definite integral, and you will have the volume.
If you don't know how to integrate those by hand, just plug in those definite integrals into wolfram alpha, and you'll be set.
Part 2: After this rotation is made, you will have an infinite number of increasing radius cylinders on the same interval in part 1 (0 to 1/3 pi). So what is the volume of a cylinder? (pi)(r)^2 (height of cylinder). We know pi, and the radius is represented by the function tan(x). And the height of the cylinder (which in theory is infinitely small) is represented by the dx in our integral. So:
A definite integral from 0 to 1/3 pi of (pi (tanx)^2 dx. Solve the definite integral, and you will have the volume.
If you don't know how to integrate those by hand, just plug in those definite integrals into wolfram alpha, and you'll be set.
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area = integral of (y dx) , x = 0 to x = pi/3
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Volume = integral of (pi y^2 dx) , from x = 0 to x =pi/3
area = integral of (y dx) , x = 0 to x = pi/3
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Volume = integral of (pi y^2 dx) , from x = 0 to x =pi/3
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π ʃ tan²x dx, between π/3 and 0
The integral of tan²x is tanx - x.
The integral of tan²x is tanx - x.