Question

find the area of the region x=2y^2 , x=4+y^2

Answer 1

you just need to subtract the integrals of the two functions

Answer 2

but first you need to figure out where they intersect

Answer 3

so set

2y^(2) = 4 + y^2
2y^(2) - y^(2)= 4
y^(2) = 4
therefore they intersect at
-2 and 2

Answer 4

I mean when y = -2 and 2

Answer 5

ok i got that part

Answer 6

so the integral is from -2 to 2

Answer 7

ok i got it now

Answer 8

thank you for the hlep

Answer 9

*help

Answer 10

You need to graph to see which is the top function and which is the bottom function

so the points of intersection are:
x = 2(2)^(2) = 8
so we have the point (-2, 8) and (-2,8) for x = 2y^(2)

now find the other two points

x = 4+ (2)^(2) = 8
thus intersection for this function is (-2,8) (-2,8)

btw I'm writing x as y and y as x

now we just need to analyze the functions by taking the derivative or just analyzing them

f(x)=2x^2
and
g(x)=4+x^2


we can tell these are parables
find the intersections and it should be easy to graph them

g(x) has y intercept at (0,4)
and no x intercept
f(x) has y intercept at 0,0
f(x) has x intercept at 0,0
thus we have the drawing