Thread: Area of a plane region

1. Area of a plane region

This is an application of integrals:
Heres the question
Find the area of the region bounded by the curves:
y^3 = x^2
and
x- 3y + 4 = 0

The answer is 27/10 square units

2. Here's the solution on a PDF file.

This is an application of integrals:
Heres the question
Find the area of the region bounded by the curves:
y^3 = x^2
and
x- 3y + 4 = 0

The answer is 27/10 square units

Hello,

first you have to calculate the intercepts of both curves:

y^3 = x^2 ===> y = (x^2)^(1/3)

x-3y+4 = 0 ===> y = (1/3)x + 4/3

Calculating the intercepts:

(x^2)^(1/3) = (1/3)x + 4/3

Raise both sides to the power of 3:

x^2 = (1/27)x^3 + (4/9)x^2 + (16/9)x + (64/27)

Multiply both sides by 27 and collect all terms at the LHS:

x^3 - 15x^2 + 48x + 64 = 0. Now some miracle occurs (not really I only noticed that you can easily tranform the term so that (x-8)² can be factored out)

x^3 - 16x^2 + 64x + x^2 - 16x + 64 = 0

x(x² - 16x + 64) + (x² - 16x + 64) = 0

(x + 1)(x² - 16x + 64) = 0 Thus you get the solutions: x = -1, x = 8

Now you know the borders of the integral.

The enclosed area is the integral of the difference of functions:

d(x) = (x²)^(1/3) - (1/3)x - (4/3)

A = int[from -1 to 8]((x²)^(1/3) - (1/3)x - (4/3))dx

A = [from -1 to 8]((3/5)x^(5/3) - (1/6)x² - (4/3)x)

Plug in the values and you'll get:

A= -(284/15) - (53/30) = -(207/10)

Could it be that there is a typo in the answer key?

EB

I've attached a diagram.

Edit: There must be a mistake in my calculations because this area isn't greater than 20 square units .... I can't find it ... sorry!

4. Yep double checked it
Thanks alot earboth & krizalid!