please can you help me,application in integration?

Question

i cn't find the answer,this is da question "find the area in the 3rd quadrant bounded by the curve x=y^2+2*y ?
thanks

another question find the area bounded by 9-x^2 and y-axis? thanks alot

Answer 1

well dear for your first question . i would show you this formula that help you how you can integrate of somthing like that.

if f(x) = x^n so ∫f(x) dx = x^ (n+1) / n+1
the other things i have to remind you is if ' a ' is a real number so we have ; ∫a dx = ax ; ∫a * x dx= a ∫x = a * x^2/x }
now we just do that;

x = y^2+2*y
∫y^2 + 2*y =
∫y^2 + ∫2*y =
y^3 / 3 + 2 (y^2)/2 , just remove ' 2 '
so we have ;
∫y^2 + 2*y .dy = y^3 / 3 + y^2 + c

now finding area;
∫y^2 + 2*y (-2,0) = [ y^3 / 3 + y^2] (-2,0)
{ u just remember first you have to put the bigger number in the formula and aas you know ' 0 > -2 ' }
[ 0 ^3 / 3 + 0^2 ] + [-2 ^3 / 3 + -2^2] = 0 + [ -8/3 + 4] = (-8 + 12) /3 =
+4/3 so the area is + 4/3

and your second question.
y = 9-x^2
∫9-x^2 dx= ∫9 -∫x^2 = 9x - x^3/3 +c
[ 9x - x^3/3 ] (-2,0) = [ 9* 0 - 0^3/3] - [ 9*-2 - (-2^3/3)]=
0 -[ -18 - (-8/3)] = +18 + 8/3 = +62/3 and its the area we were looking for.

Well darling i hope i could help you.
good luck .

Answer 2

Integrate from -2 to 0 the function (y^2 + 2y) dy


Answer: -4/3

Answer 3

graph f(y)=y^2+2y: left function
would be bounded by x=0 or g(y)=0: right function

A= ~(0, -2) g(y)-f(y) dy
A= ~(0, -2) 0-(y^2+2y)dy
A= ~(0, -2) -y^2-2y dy
A= [-(1/3)y^3-y^2] (0, -2)
A= [-(1/3)0^3-0^2] - [-(1/3)(-2)^3-(-2)^2]
A= 0 - [8/3-4]
A= 4/3