Find the volume obtained by rotating the area (calculus question)?

hi, i would like to find out how to solve this problem. i am asking again as when i first asked i didn't get many answers and the ones i did get were incorrect anyway...





i would like to find out how to solve this problem. i would like to find the volume obtained by rotating the area between


y = 6 - ( x-1 )^ 4 and y = 5 around the y axis





i would like to find the exact answer (no decimals)





thanks for any help. it is very much appreciated

Find the volume obtained by rotating the area (calculus question)?
1st, we have to find where the area begins and ends.


Let us find the coincident points of the 2 functions:


6 - ( x-1 )^ 4 = 5


1 = (x-1)^4


So, x-1 can be either 1 or -1 andd then x=0 or x=2


Then we get to the calculus part:


We know the volume of a solid of revolution is INTEGRAL(pi·[f(x)]²dx), if it is around the x axis, and INTEGRAL(2pi·x·f(x)dx) if is the y axis.


Calculating the volumes, for each curve:


Volume of 6-(x-1)^4 is 116PI/5 and the other is 20PI


which results 16PI/5
Reply:You didnt get many answers because people arent gona do your work for you. BTW, i cant help you cause im doing gr 11 math, learning about complex numbers.
Reply:y = 6 - ( x-1 )^ 4





substituting y = 5, we obtain the limit, x = 0 %26amp; 2





simplifying y = 6 - ( x-1 )^ 4 we get





y = - x^4 + 4·x^3 - 6·x^2 + 4·x + 5





since





Area = ∫dA





∫dA = ∫ydx





∫dA = ∫(- x^4 + 4·x^3 - 6·x^2 + 4·x + 5)dx





A = - x^5/5 + x^4 - 2·x^3 + 2·x^2 + 5·x





substituting the limit will give you





A = 58/5





by 2nd proposition of pappus





vol = A * 2R, where R = is equal to the distance of the centroid from the rotating line. R = (1/2) the distance between pt(0,5) %26amp; (2,5), in your case. R = 1





V = 2(58/5)(1)





V = 116·/5 cu. units

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