# Thread: Volume (Integral)

1. ## Volume (Integral)

Solution:
y = -2
(-2)^2 = x
x = 4
y^2 = +- 2
the point of intersection should be (0,0) and (4,2)

now im using the cylindrical method in obtaining the volume
$V = 2\pi\int_{0}^{4} R dA$
where dA = (y2-y1)dx
for
y2 = sqrt(x)
y1 = x
and
radius be r = x+2 since the revolved line is at the 2nd quadrant and x in the 1st quadrant and it needs to be added

$V = 2\pi\int_{0}^{4} (x+2) (y_2 - y_1) dx$
$V = 2\pi\int_{0}^{4} (x+2) (\sqrt{x} - x) dx$
V = -(416/15)pi ????

2. ## Re: Volume (Integral)

really hard to see your tiny image

3. ## Re: Volume (Integral)

Originally Posted by skeeter
really hard to see your tiny image
im done editing
hope someone replies

edit
wut was this sarcasm?
i posted as an image so that other students in my class will not google this question
i want to resize the pic hoewer something might go wrong with the image and im in a hurry

4. ## Re: Volume (Integral)

Originally Posted by MathMinors
im done editing
hope someone replies

edit
wut was this sarcasm?
i posted as an image so that other students in my class will not google this question
i want to resize the pic hoewer something might go wrong with the image and im in a hurry
no sarcasm at all ... the image is so small that one cannot read it.

why is it so important to keep others from seeing the question?

5. ## Re: Volume (Integral)

Originally Posted by skeeter
no sarcasm at all ... the image is so small that one cannot read it.

why is it so important to keep others from seeing the question?
Probably so that his/her lecturer doesn't find out that he/she's seeking help for this assessment...

6. ## Re: Volume (Integral)

Originally Posted by Prove It
Probably so that his/her lecturer doesn't find out that he/she's seeking help for this assessment...
Maybe, maybe not ... in any case, something's not right. Thread closed.