area bounded by curves

[Guest]

This section always troubles me because i never know whether to put the strip horizontal or vertical. These 2 questions are really troubling me...


Determine the area bounded by curves...

1. y^2=x^3 and x=4
2. y^2 =2x and x^2=2y

Any help would be greatly appreciated!

 

[stapel]

never know whether to put the strip horizontal or vertical.

It's a matter of choice.

Determine the area bounded by curves:
1. y^2=x^3 and x=4

Since "x = 4" is a vertical line, then probably horizontal strips would be best. So solve "y² = x³" for "x=".

2. y^2 =2x and x^2=2y

This one could go either way. If you want to go vertical, then use "y = ± sqrt[2x] and y = x²/2". Otherwise, solve for "x=" on each equation.

Eliz.

 

[Guest]

I figured, both ways would work and i tried both ways for Q1, but the problem is when i go to factor, it doesnt work, it just doesnt factor. For example...

when i solved for x for both y^2=2x and x^2=2y in question 1, i ended up with y^4 =8y , but that factors to x(x-2)(x^2+2x+4) which doesnt give me any limits. How would i do the question?

 

[stapel]

i ended up with y^4 =8y , but that factors to x(x-2)(x^2+2x+4)...

Do you perhaps mean "y⁴ = 8y gives me y(y - 2)(y² + 2y + 4) = 0"...? And why can you not solve this for two zeroes...?

Eliz.

 

[galactus]

For problem #2:

Have you graphed. If you do, it helps a lot.

\(\displaystyle y^{2}=2x\) and \(\displaystyle x^{2}=2y\)

Solving for y=

\(\displaystyle y=\sqrt{2x}\) and \(\displaystyle y=\frac{x^{2}}{2}\)

\(\displaystyle sqrt{2x}=\frac{x^{2}}{2}\), x=0 or 2.

\(\displaystyle \int_{0}^{2}({sqrt{2x}-\frac{x^{2}}{2}})dx\)