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Page history last edited by PBworks 14 years, 8 months ago

Question 1


The use of a graphing calculator is allowed.


Let R be the region in the first quadrant bounded by the graph of y = 4 - x2/3, the x-axis and the y-axis.


(a) Find the area of region R.


(b) Find the volume of the solid generated when R is rotated around the x-axis.


(c) The vertical line x = k divides the region R into two regions so that when these regions are rotated about the x-axis, they generate solids with equal volumes. Find the value of k.




y=4-x^(2/3) and y=0

4-x^(2/3)=0 (find the intersection between the graph and x-axis)

x=8 Is where the two lines intersect and enclose the Area of region R.

(Σ takes the place of the integral sign and the numbers after it are the (a,b) of the integral.)

Σ(0,8)4-x^(2/3) dx

=(4x-(3/5)x^(5/3))(0,8) (the area of R is the integral of the function 4-x^(2/3) from (0,8))

A= [4(8) -(3/5)(8^(5/3))]-[4(0) -(3/5)(0^(5/3))] (Substitute 8 and 0 into the function 4-x^(2/3) to solve for the Area of region R)

A= [12.8] -0





=91.8951 (the volume of the solid πr^2 and radius is the function,and just use calculator to calculate the value is 91.8951)


c) x=K divides the region into two regions


V=πΣ(0,k) (4-x^(2/3))^2 dx=91.8951/2 (The volume is divided by 2 in order to solve for the x value of k which divides the volume generated by rotating region R into two equal volumes.)

[πΣ(0,k)(4-x^(2/3))^2]/π = 45.94755/π (Divide both sides by π)

Σ(0,k)(4-x^(2/3))^2=14.6286 Antidifferentiate the integral)

(16x-(24/5)x^(5/3)+ (9/13)x^(13/9))(0,K)=14.6286 (The next step involves substitution)

16K-(24/5)k^(5/3)+(9/13)k^(13/9)=14.6286 (Solve for k)


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