Area of rectangle - Area under a curve (1 Viewer)

[MrBrightside]

Hi, this question, i have done all parts, but at part (iv) I cannot not define this logic.

I went from the focus point (0,5) to the vertex (0,3). a = 2

therefore, when y = 5

x^2 = 8(5-3)
= 16
x = +-4

Therefore, the area of the rectangle (from my way) I did 8*2, as x = -4 and 4, therefore the base of the rectangle is 8 units. and then I went from the vertex to the focus point, which is 5-3 = 2 units.

Then I just did area under y = -(x^2 / 8) + 3 using the limits 4 to 0 multiplied by 2.

I got 16 - 2[28/3]

But why doesn't this give the correct answer?

 

[SpiralFlex]

For points of intersection,





Rearrange,





Since it's an even function,





with limits 0, 4.








Note: Your error was when you used your rectangle box, you were trying to integrate and restrict it to the rectangle. However when you integrate, you are actually finding the area bounded by the axis.


I drew a pretty diagram for you.


Here is what you were trying to do.




Here is what you ACTUALLY did.

 

[MrBrightside]

For points of intersection,





Rearrange,





Since it's an even function,





with limits 0, 4.








Note: Your error was when you used your rectangle box, you were trying to integrate and restrict it to the rectangle. However when you integrate, you are actually finding the area bounded by the axis.

Oh okay, thanks.

The book's ans did it differently though, but it makes sense, becasue they went right to the x-axis getting the rectangle's dimensions to 8*5 = 40

but wtf, the book doesn't put a minus in front of (x^2/8) :/

 

[accountant]

Spiril there is a question VCE, why aren't you answering it for?

 

[SpiralFlex]

Spiril there is a question VCE, why aren't you answering it for?

I did in PM. ^_^

 

[MrBrightside]

For points of intersection,





Rearrange,





Since it's an even function,





with limits 0, 4.








Note: Your error was when you used your rectangle box, you were trying to integrate and restrict it to the rectangle. However when you integrate, you are actually finding the area bounded by the axis.


I drew a pretty diagram for you.


Here is what you were trying to do.




Here is what you ACTUALLY did.

Hmmm It's funny, cause I remember doing a question similar to this in class and it worked by doing it the first way. : / so trippy shit. It's easy, but trippy cause I didn't know the area gets extended all the way to the x-axis. Thanks for this wonderful piece of advice.

And to this guy that makes new accounts every time i post something, I will try my best to get a good mark.

What graphics package is that for the graphs?

 

[SpiralFlex]

Hmmm It's funny, cause I remember doing a question similar to this in class and it worked by doing it the first way. : / so trippy shit. It's easy, but trippy cause I didn't know the area gets extended all the way to the x-axis. Thanks for this wonderful piece of advice.

And to this guy that makes new accounts every time i post something, I will try my best to get a good mark.

What graphics package is that for the graphs?

Geogebra.

 

[MrBrightside]

Awesome spiral, thanks I did it two different ways and got them both right! guess I just needed some sleep to refocus. I made one silly algebraic error, because I skipped writing 1 line of working ><, so not doing that again, I'm going to write every line of working in the test. It's too easy to fall down over a little algebraic error.