Disc Method Finding Volume Trig Function

Chaoticoli
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Homework Statement


Consider the region of the x-y plane between the line y=7, the curve y=3sin(x)+4 , and for
-pi/2<x<3pi/2

Find the volume of the solid generated by revolving this region about the line y=7.

I know that for this problem, I will be using the disk method (as the title states). I drew the graph and I find that there are two areas and since the graph is symmetric, the two areas are equal.

A picture of the graph is here: http://www.wolframalpha.com/input/?i=+the+curve+7=3*sin(x)+4+between+-pi/2<x<3pi/2

I know I must find the area of a slice. I am just not exactly sure how to find it. I know the equation for the area is A(x) = pi*r(x)^2

And then, you need to find the limits of integration and with those limits, the volume will be:

Volume = ∫ from a to b of pi*A(x)

Any help would be greatly appreciated ! :)
 
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Obviously, I just need to find out what r(x) is and figure out if I should integrate with respect to x or y.
 
Chaoticoli said:
since the graph is symmetric, the two areas are equal.
Not exactly. The area of interest is above the curve, so they're not so much equal as, in a sense, complementary.
Anyway, finding the area of regions in this graph won't help you.
I know I must find the area of a slice. I am just not exactly sure how to find it. I know the equation for the area is A(x) = pi*r(x)^2
Yes.
Can you visualise the rotation about y=7?
r(x) is the distance between two y values at x. What two y values?
 
Yes. I visualize it being rotated, forming discs about the line y=7.

Is it the distance between y=3sin(x)+4 and y=7?

(3sin(x) + 4) - 7 = 3sin(x) -3 = r(x) ??
 
Actually, I just figured it out :). I got 27pi^2 after integrating from -pi/2 to 3pi/2 of (3sin(x)-3)^2 dx
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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