# Volume of a Solid by Revolving

• Jul 28th 2009, 07:23 AM
Brazuca
Volume of a Solid by Revolving

The answers don't make sense with what I have learned.
for A) about the x-axis shouldn't it be (Left - Right)

Shouldn't it have been: PI (antiderivative) of [(0)^2 - (x^3)^2] dx

Why did they put (1)^2 instead of 0 when 0 is on the left.

I understand part B, but A doesn't make sense to me.
• Jul 28th 2009, 07:43 AM
chengbin
The question stated y=1, which means it is capped off to y=1

The integral is setup as $\int_a^b \pi[f(x)^2-g(x)^2]dx$

Where f(x) is the graph "above" of g(x).

1 is above the graph $y=x^3$ from the domain $0, therefore $1^2$ is written first.
• Jul 28th 2009, 07:46 AM
Brazuca
I don't understand why it would be like that for $\int_a^b \pi[f(x)-g(x)]dx$

Shouldn't it be Left - Right when it is like that?
• Jul 28th 2009, 09:07 AM
Brazuca
When doing it "about the X-axis" shouldn't it look like this
$\int_a^b \pi[Left-Right]dx$

Also why does the Y = 1 have any effect on that since it is dx and not dy?
• Jul 28th 2009, 09:34 AM
skeeter
Quote:

Originally Posted by Brazuca
When doing it "about the X-axis" shouldn't it look like this
$\int_a^b \pi[Left-Right]dx$

Also why does the Y = 1 have any effect on that since it is dx and not dy?

about the x-axis, the method of washers volume is

$V = \pi \int_a^b [f(x)]^2 - [g(x)]^2 \, dx$

where f(x) is the upper function and g(x) is the lower function.

about the y-axis, the method of washers volume is

$V = \pi \int_c^d [f(y)]^2 - [g(y)]^2 \, dy$

where f(y) is the right function and g(y) is the left function.
• Jul 28th 2009, 10:15 AM
Brazuca
OHH That is so much easier than the way I learned it.