- #1
Pearce_09
- 74
- 0
Hello,
Consider the integral:
[tex] \int_R x^2 + y^2 dA [/tex]
with the two graphs 2x-y = 0 and [tex] x^2 [/tex] - y = 0
therefore y = [tex] x^2 [/tex] and y = 2x are the two functions
and the point of intersection is at (0,0) and (2,4)
therefore
[tex] \int { \int x^2 + y^2 dx } dy [/tex]
(a - is top point of the integral and b - is the bottom)
therfor the domain for the first integral (dx) is b = y/2 and a = [tex] y^1^/^2 [/tex]
and for the second integral (dy) is b= 0 and a = 4
but when i switch the order to """"" dy dx... i get a different #.
therefore my new a,b for the integrals are
for the first integral (dy) b = [tex] x^2 [/tex] a = 2x
for the second integral (dx) b = 0 a = 2
is my change of order correct or did i do somthing wrong??
Consider the integral:
[tex] \int_R x^2 + y^2 dA [/tex]
with the two graphs 2x-y = 0 and [tex] x^2 [/tex] - y = 0
therefore y = [tex] x^2 [/tex] and y = 2x are the two functions
and the point of intersection is at (0,0) and (2,4)
therefore
[tex] \int { \int x^2 + y^2 dx } dy [/tex]
(a - is top point of the integral and b - is the bottom)
therfor the domain for the first integral (dx) is b = y/2 and a = [tex] y^1^/^2 [/tex]
and for the second integral (dy) is b= 0 and a = 4
but when i switch the order to """"" dy dx... i get a different #.
therefore my new a,b for the integrals are
for the first integral (dy) b = [tex] x^2 [/tex] a = 2x
for the second integral (dx) b = 0 a = 2
is my change of order correct or did i do somthing wrong??