Find the area bounded by y=8-x^2 and y=x^2.
- print Print
- list Cite
Expert Answers
calendarEducator since 2010
write12,544 answers
starTop subjects are Math, Science, and Business
We first determine the points where the curves y = 8 - x^2 and y = x^2, meet.
8 - x^2 = x^2
=> x^2 = 4
=> x = 2 , x = -2
Now...
(The entire section contains 74 words.)
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Related Questions
- Calculate area bounded by the curve y=cosx/(sin^2x-4), x axis and the lines x = 0 and x = pi/2 .
- 1 Educator Answer
- What is the area of the region enclosed by the curves y=lnx and y=ln^2x ?
- 1 Educator Answer
- What is the area of the region enclosed between the curves y=x^2-2x+2 and -x^2+6 ?
- 1 Educator Answer
- 1) What are the areas bounded by the lines and the x axis? In each case x>0 a) y = 3/xb)...
- 1 Educator Answer
- What is the area of the region bounded by the curve y=square root(x-1), y axis and y=1 to y=5 ?
- 1 Educator Answer
First, we'll have to identify the limits of integration. For this rason, we'll determine the intercepting points of the graphs.
Since y = x^2 and y = 8-x^2, we'll get:
x^2 = 8-x^2
We'll move all terms to one side:
x^2 - 8 + x^2 = 0
2x^2 - 8 = 0
We'll divide by 2:
x^2 - 4 = 0
x^2 = 4
x1 = 2 and x2 = -2
Over the interval [-2 ; 2] the graph 8-x^2 >= x^2
The area of the region is the definite integral of the function 8 - x^2 - x^2, for x = -2 to x = 2.
A = Int (8 - x^2 - x^2)dx
A = Int (8 - 2x^2)dx
A = Int 8dx - Int 2x^2dx
A = 8x - 2x^3/3, for x = -2 to x = 2
A = 8(2 + 2) - (2/3)(8 + 8)
A = 32 - 32/3
A = (96-32)/3
A = 64/3 square units.
The area is A = 64/3 square units.
Student Answers