Please help me with this Area between Curves question, please provide a video

Question

Sketch the region enclosed by a given curve. Decide whether to intehrate with x and y

y= 8x^2 ,y^2=1/8x

I did the calculation i know the limit is 1/8 and 0 but I just dont know how to compute it correctly i keep getting an impossible long negative number in my calculator so i know i did something wrong.

Answer 1

Well the way I approach it is to take the limits and integrate the top - bottom function.

Integrate the limit from 0 to 1/8 of (x/8)^1/2- 8x^2.

Integrate (x/8)^1/2- 8x^2

(2/3*(1/8)^.5)x^3/2-(8/3)x^3

From there inputting your limits of 1/8 and 0 which when you input 0 should get you 0 for that section, so you only need to worry about the calculation when inputting your limit of 1/8 which gets about 1/192. Apologies that I can’t show the graph answering on my phone.

Answer 2

To find the intersection point substitute "8x²" for y in the second equation:

y² = 1/8x

(8x²)² = 1/8x

64x⁴ = 1/8x

64x⁴ - 1/8x = 0

x(64x³ - 1/8) = 0

x = 0 and 64x³ - 1/8 = 0 so x³ = 1/512 or x = 1/8

Plugging this into either equation yields (for x = 0) y = 0 and (for x = 1/8) y = 1/8 so the intersection points are (0, 0) and (1/8, 1/8)

Graphing yields:

I'm going to choose to integrate from x = 0 to x = 1/8.

The area between the curves is ∫y_blue dx - ∫y_red dx (from x = 0 to x = 1/8

y_blue:

y² = 1/8x

y = √(1/8x)

y = (1/8x)^1/2

y = √2/4x^1/2

y_red:

y = 8x²

Area = ₀∫^1/8√2/4x^1/2dx - ₀∫^1/88x²dx

Area = [(√2/4)(2/3)x^3/2 - 8/3x³] evaluated between 0 and 1/8

Arae = √2/6(1/8)^3/2 - 8/3(1/512)

Area = 1/96 - 1/192

Area = 2/192 - 1/192 = 1/192