Set up the integrals to evaluate the area of the planar region bounded by (y^2)-8y+13, y-x= 9, y = 2 and y = 7   by horizontal slicing, by vertical slicing. Evaluate only for the method...

Set up the integrals to evaluate the area of the planar region bounded by (y^2)-8y+13, y-x= 9, y = 2 and y = 7

 

by horizontal slicing, by vertical slicing.

Evaluate only for the method that yields the fewer number of integrals.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to check what is the upper graph and what is the lower graph, over the interval `[2,7]`  such that:

`y = 2 => 2^2 - 8*2 + 13 = 1`

`y = 2 => 2 - 9 = -7`

`y = 7 => 7^2 - 8*7 + 13 = 6`

`y = 7 => 7 - 9 = -2`

Notice that the upper curve is `x = y^2 - 8y + 13`  and the lower curve is `x = y - 9` .

You need to evaluate the definite integral such that:

`int_2^7 (y^2 - 8y + 13 - (y - 9)) dy = int_2^7 (y^2 - 9y + 22) dy`

You need to use the property of linearity of integral such that:

`int_2^7 (y^2 - 9y + 22) dy = int_2^7 (y^2)dy - int_2^7 (9y) dy+ int_2^7 (22) dy`

`int_2^7 (y^2 - 9y + 22) dy = (y^3/3 - 9y^2/2 + 22y)|_2^7`

Using the fundamental theorem of calculus yields:

`int_2^7 (y^2 - 9y + 22) dy = 7^3/3 - 9*7^2/2 + 22*7 - 2^3/3 + 9*2^2/2 - 22*2`

`int_2^7 (y^2 - 9y + 22) dy = 343/3 - 441/2 + 154 - 8/3 + 18 - 44`

`int_2^7 (y^2 - 9y + 22) dy = 335/3 + 128 - 220.5`

`int_2^7 (y^2 - 9y + 22) dy = 335/3- 92.5`

`int_2^7 (y^2 - 9y + 22) dy~~ 19.16`

Hence, evaluating the definite integral yields `int_2^7 (y^2 - 9y + 22) dy ~~ 19.16.`

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