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Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. 

 

2x+y^2=8 , x=y

 

got stuck on this one

 Apr 30, 2019
 #1
avatar+101870 
+2

2x + y^2  = 8  ,   x  =  y

 

Find the intersection points    by subbing the  second equation into the first

 

2x + x^2 = 8      rearrange  as

 

x^2 + 2x - 8  = 0       factor

 

(x - 2) ( x + 4)  = 8     set both factors to 0  and solve for x  and we have that  x = 2  and x = - 4

 

Here's the graph : https://www.desmos.com/calculator/jdelptugab

 

Then the intersection points are  ( -4, -4)  and (2, 2)

 

Because of the presence of the y^2 term....this might be easier to integrate in terms of y

 

Notice that we can write   2x + y^2 = 8   as    x = [ 8 - y^2] / 2  = 4 - (1/2)y^2.....and this is the function on the "right"  and y = x is the function on the "left"

 

So we have

 

 2

 ∫    [ 4 - (1/2)y^2] - y    dy    = 

-4

 

                                         2

[ 4y - (1/6)y^3  - (1/2)y^2 ]   =    

                                        -4

 

[ 4(2) - (1/6)(2^3) - (1/2)2^2 ]  - [ 4(-4) - (1/6)(-4)^3 - (1/2)(-4)^2 ]  =

 

[ 8 - 4/3 - 2 ]  - [ -16 + 32/3 - 8 ]   =  

 

14/3   - [- 40/3]  =  

 

54/3  = 

 

18 units^2 

 

cool cool cool

 Apr 30, 2019
 #2
avatar+322 
+1

thank you

Ruublrr  Apr 30, 2019
 #3
avatar+101870 
+1

OK, man  !!!!!

 

 

Glad to help......

 

cool cool cool

CPhill  Apr 30, 2019

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