Multiple parts

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Lizzie

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Ok, these pieces have to do with the area bound by the equation y=x<sup>2</sup>, the y-axis, and y=4.

A) The area of this region is?

Ok, how do I do this when the graphed area is up against the y-axis instead of the x-axis?

B) The volume obtained when the region is rotated about the x-axis is?

If someone could explain what the heck this is talking about and how to do it, I could probably figure out C as well.

C) The volume obtained when the region is rotated about the y-axis is?

Remember, I am not asking for the solution, please don't think otherwise. Thank you very much!
 
A) To find the area "between", you subtract the lower from the upper. Before, the lower was the x-axis, also known as "y = 0". Since subtracting zero is rather boring, this was omitted.

Now, however, the upper is "y = 4" and the lower is "y = x<sup>2</sup>", so you will need to account for both the upper and the lower in your integral.

B&C) Has your class covered "by shells" and/or "by washers" yet? If not, we can try to find some links because, without pictures, there's about no chance you'd understand the explanation....

Eliz.
 
stapel said:
A) To find the area "between", you subtract the lower from the upper. Before, the lower was the x-axis, also known as "y = 0". Since subtracting zero is rather boring, this was omitted.

Now, however, the upper is "y = 4" and the lower is "y = x<sup>2</sup>", so you will need to account for both the upper and the lower in your integral.
would it be something like taking the integral of y=x^2-4? (probably not, I have to recheck my notes)
B&C) Has your class covered "by shells" and/or "by washers" yet? If not, we can try to find some links because, without pictures, there's about no chance you'd understand the explanation....
Uhm, nope, but I know a website that explains that a little bit. At least, I know it gives some formulas or something.
Eliz.
Click to expand...
 
Man, when I graduate from college, I am most definatley going to become a tutor on here during my free time.
 
stapel said:
A) To find the area "between", you subtract the lower from the upper....the upper is "y = 4" and the lower is "y = x<sup>2</sup>"....
Click to expand...
Lizzie said:
would it be something like taking the integral of y=x^2-4?
Click to expand...
Which is the upper function (the "top" of the region) and which is the lower (the "bottom")?

Subtract the bottom from the top.

Eliz.
 
Oh duh, we just did that recently, didn't we? I think I remember it now. Thanks for reminding me. I swear, I must just be freaking out since my final is so close because I am completely missing so many small things anymore.
 
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