How do you do the area of bounded y2 equals 9xand2x equals 9y?

Answer 1

The basic steps you must follow are:

1. Simplify the two equations to be functions of y.
2. Find the intersections of the two functions.
3. Integrate the difference of the two functions on between the two intersection points.

Here are the steps for this specific problem:

Simplifying the equations:

Simplify y2=9x to be a function of y.

y=±√(9x)

y=±3√(x)

(Keep the ± for now.)

Simplify 2x=9y to be a function of y

(2/9)x=y

y=2/9x

Finding the intersection of the two functions:

±3√(x)=y=2/9x

±3√(x)=2/9x

±27/2√(x)=x

x=0

±27/2=√(x)

729/4=x

x=729/4

(Now, since neither of the solutions is negative, the ± may be removed from the first function.)

Integrating the difference of the two functions:

Let A=area bounded by the two curves.

Let y1=3√(x), the first function.

Let y2=2/9x, the second function.

A=∫(y1-y2)dx

(y2 is subtracted from y1 because it is the lower function on that interval. Graph the two functions; it makes sense.)

A=∫(3√(x)-2/9x)dx on [0, 729/4]

A=∫(3x1/2-2/9x)dx on [0, 729/4]

A=2x3/2-1/9x2+C on [0, 729/4]

Now, find the particular solution on the interval [0, 729/4]

A=[2(729/4)3/2-1/9(729/4)2+C]-[2(0)3/2-1/9(0)2+C]

A=19683/4-59049/16+C-C

A=19683/4-59049/16=19683/16=1,230.1875

The area of the region bounded by y2=9x and 2x=9y is 1,230.1875 units2.