Find the area of the region R bounded by y= x^{6}/3- 4, the x-axis, x= -1, and x= 1.

Calculus
Tutor: None Selected Time limit: 1 Day

Find the area of the region bounded by , the -axis, , and .  How would I solve this?

Apr 25th, 2015

given y=x^{6}/3-4 between x=-1 and x=1  using integral calculus method

integrating y with respect to X gives area (A),when we substitute boundary conditions

the integral will be A= [x^{7}/21-4x+c] where c=arbitrary constant, A= area in square units

but taking boundary conditions x= -1 and 1 and substituting into equation,

A= [(1/21-4)-(1/21+4)]

area = 8  square units ( N/B negative sign is assumed for area)

Apr 25th, 2015

used integral calculus method to solve it. arbitrary constant will cancel for an homogenous integrals due to boundary given.

Apr 25th, 2015

How do you account the vertical y-axis values where the minimum is -4 to x-axis?

Apr 25th, 2015

for such you will only take boundary conditions as x=-4 and x=0, in other wards you consider from 0 down to negative integers. otherwise your equation would not make sense if you substitute directly negative values of x with positive ones.

Apr 25th, 2015

basically you're saying to take the integration (1/3)*x^6 - 4 w/ respect of x in interval from 0 to 1 multiplied by 2.  How is the X-Axis boundary accounted for?

Apr 25th, 2015

Because for whatever the reason, I did exactly that, and the integration of x = (3*y + 12)^(1/6) with respect of y w/ interval from -4 to 0.  I have two different answers.  w/ respect of y-axis it is 5.18 and w/ respect of x-axis it is -7.905.  Why am I getting two different values altogether?

Apr 25th, 2015

good argument and reasoning

simple, initially your equation was function of x right? that is y(x) but you decided to express x in terms of y giving you x as function of y i.e x(y) right and you got different answers well, 

if you express in terms of y, then boundary conditions will never be x=-4 or 0, you use values of y on given point.

example, points (-4,y1) and (0,y2) so your limits will be y1 and y2

thank you



Apr 26th, 2015

N/B given boundary values as real values of x and function in terms of x then integrate with respect to x

and if you decide otherwise by expressing function in terms of y, then use boundary values as real values of y at same given point

Apr 26th, 2015

N/B given boundary values as real values of x and function in terms of x then integrate with respect to x

and if you decide otherwise by expressing function in terms of y, then use boundary values as real values of y at same given point

Apr 26th, 2015

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