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Mathematics 13 Online
OpenStudy (anonymous):

Help finding volume of solid

OpenStudy (anonymous):

OpenStudy (anonymous):

@satellite73

OpenStudy (anonymous):

@zepdrix

OpenStudy (anonymous):

draw

OpenStudy (anonymous):

desmos?

OpenStudy (anonymous):

you need to draw the thing first.

OpenStudy (anonymous):

https://www.desmos.com/calculator/abc9opgrzj

OpenStudy (anonymous):

okay, so now, do you want to use disk method or what?

OpenStudy (anonymous):

yeah disk

OpenStudy (anonymous):

\[ V = \int_{-\pi/2}^{3\pi/2}\pi r^2(x)\;dx \]

OpenStudy (anonymous):

What do you think \(r(x)\) would be in this case?

OpenStudy (anonymous):

That's what was giving me trouble. Do I subtract the 4 sin(x)+2 and 6? I dont think that's right. How do I use double angel formulas with this problem?

OpenStudy (anonymous):

uh, it's just \(6-4\sin(x)+2\)

OpenStudy (anonymous):

you have to square it.

OpenStudy (anonymous):

Right I understand that part. I was just confused because the program said to use the double angel trig formulas...

OpenStudy (anonymous):

i think I got it from here, thanks

OpenStudy (anonymous):

OpenStudy (anonymous):

you will have to deal with a squared sin term.

OpenStudy (anonymous):

which is dealt with using half angle, or even double angle.

OpenStudy (anonymous):

\[\int\limits_{?}^{?}16 \sin^2 (x) dx= 8x-4\sin(2) ???\]

OpenStudy (anonymous):

ive never used double angle before

OpenStudy (anonymous):

\[ \sin^2(x) = \frac{1-\cos(2x)}{2} \]

OpenStudy (anonymous):

It's really the half angle formula in this case.

OpenStudy (anonymous):

8(1-cos(2x))

OpenStudy (anonymous):

@Hamoody1996

OpenStudy (anonymous):

are you still stuck?

OpenStudy (anonymous):

is it : pi(8(1-cos(2x)-64cos(x)+64x

OpenStudy (anonymous):

Is that the integrand you are getting?

OpenStudy (anonymous):

that's the indefinite integral

OpenStudy (anonymous):

why do you have two \(\cos\) in there?

OpenStudy (anonymous):

48pi^2 is the answer. I used 4sin(x)-4

OpenStudy (anonymous):

as r(x)

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