Thread: Calc II problem

I am trying to approximate the the area of a curve. I can easily do this, it's just that when I look at a graph of this new problem I do not see it a "shaded region...

Re:

Replace x with y and y with x so you just tilt the graphs by 90 degree. And the area stays the same.

I am just trying to see which equation is on top so I know which one to put first and second into my integral.

x = 5y^2 and x = 9 +4y^2

Thus,

5y^2 = 9 + 4y^2

y^2 = 9

y = +3 (ignoring the bottom point)

That is where they intersect.
And chose a point above that and below that and compare values.

So in these problem you are give y in terms of x. Opposite of what I am used to dealing with. Can you still solve for y= if you are given this type of problem?

-qbkr21

You can not find the area of a curve. You can only find the area of a region.

Originally Posted by ThePerfectHacker

Replace x with y and y with x so you just tilt the graphs by 90 degree. And the area stays the same.

ThePerfectHacker will this rule work all of time? I mean any time you are given a problem in terms of y instead of x can you just switch the variables and turn in 90 degrees to the right?

-qbkr21

Originally Posted by qbkr21

ThePerfectHacker will this rule work all of time? I mean any time you are given a problem in terms of y instead of x can you just switch the variables and turn in 90 degrees to the right?

-qbkr21

Yes.

All you are doing is turning you head 90 degrees. That is what engineering professors like to say.