# Volume

• Jan 12th 2010, 02:55 PM
CarDoor
Volume
I'm having a lot of problems with these questions.

Volume of the region enclosed by y^2=4x and y=x about the line x=4

I can graph it, but I can't set up the integral.

Another one is "The base of a solid is a the region enclosed between the graphs of y=sinx and y=-sinx from x=0 to x=pi. Each cross section perpendicular to the x-axis is a semicircle with diameter connecting the two graphs. Find the volume."

I just don't know what to do with the information it gives me in the second sentence.
• Jan 12th 2010, 03:11 PM
skeeter
Quote:

Originally Posted by CarDoor
I'm having a lot of problems with these questions.

Volume of the region enclosed by y^2=4x and y=x about the line x=4

I can graph it, but I can't set up the integral.

Another one is "The base of a solid is a the region enclosed between the graphs of y=sinx and y=-sinx from x=0 to x=pi. Each cross section perpendicular to the x-axis is a semicircle with diameter connecting the two graphs. Find the volume."

I just don't know what to do with the information it gives me in the second sentence.

washers w/r to y about the line x = 4 ...

$\displaystyle V = \pi \int_0^4 \left(4- \frac{y^2}{4}\right)^2 - (4-y)^2 \, dy$

using cylindrical shells w/r to x about x = 4 ...

$\displaystyle V = 2\pi \int_0^4 (4-x)(2\sqrt{x} - x) \, dx$

for the second problem, the diameter of a semicircular cross section is the vertical distance from $\displaystyle y = \sin{x}$ to $\displaystyle y = -\sin{x}$ , therefore, $\displaystyle d = 2\sin{x}$ ... so, the radius of a semi-circular cross section is $\displaystyle r = \sin{x}$

cross-sectional area = $\displaystyle \frac{\pi}{2}r^2 = \frac{\pi}{2}\sin^2{x}$

$\displaystyle V = \int_0^{\pi} \frac{\pi}{2}\sin^2{x} \, dx$
• Jan 12th 2010, 03:18 PM
galactus
Quote:

"The base of a solid is a the region enclosed between the graphs of y=sinx and y=-sinx from x=0 to x=pi. Each cross section perpendicular to the x-axis is a semicircle with diameter connecting the two graphs. Find the volume."

The region between the functions can be denoted by

$\displaystyle y=sin(x)-(-sin(x))=2sin(x)$

This is the diameter of the semicircle. Can you picture them stacked up in

the region with the base of the semicircles being between sin(x) and -sin(x)?.

The area of a semicircle is $\displaystyle \frac{{\pi}y^{2}}{2}$

The radius is $\displaystyle \frac{2sin(x)}{2}=sin(x)$

$\displaystyle \frac{\pi}{2}\int_{0}^{\pi}sin^{2}(x)dx$
• Jan 12th 2010, 04:30 PM
CarDoor
Thanks for the help.

I still have a question though. Why was is 4-y but not 4+y? By looking at the graphing how can you tell?
• Jan 12th 2010, 04:33 PM
skeeter
Quote:

Originally Posted by CarDoor
Thanks for the help.

I still have a question though. Why was is 4-y but not 4+y? By looking at the graphing how can you tell?

what is the horizontal distance from the line x = 4 to x = y ?
• Jan 12th 2010, 04:39 PM
CarDoor
Alright. I think I got it. Thanks again for the help.
• Jan 13th 2010, 05:13 PM
CarDoor
Okay. Maybe I don't get it. What if the region was on the right of x=4? Would it be y-4?
• Jan 13th 2010, 05:42 PM
skeeter
Quote:

Originally Posted by CarDoor
Okay. Maybe I don't get it. What if the region was on the right of x=4? Would it be y-4?

right curve - left curve
• Jan 13th 2010, 06:02 PM
CarDoor
Alright thanks. I think I'm finally getting comfortable with this.