# Homework Help: Trigonometric Inegrals: Volume help

1. Feb 13, 2010

### Slimsta

1. The problem statement, all variables and given/known data
Let R be the region enclosed by the curves y=6sin2x and y=6 over the segment [ π/2,3π/2]. Find the volume of the solid that is obtained by rotating R about the line y=−1.1.

2. Relevant equations
$$\displaystyle \Large pi * \int _a^c x((r+yt)^2 - (r+yb)^2)dx$$

3. The attempt at a solution
$$\displaystyle \Large pi * \int _{pi/2}^{3pi/2} x((-1.1+6)^2 - (-1.1+6sin^2x)^2)dx$$
==> $$\displaystyle \Large pi * \int _{pi/2}^{3pi/2} 24.01x - (-1.1+6sin^2x)^2 xdx$$

==> $$\displaystyle \Large pi [* \int _{pi/2}^{3pi/2} 24.01x dx - \int _{pi/2}^{3pi/2} (-1.1+6sin^2x)^2 x dx ]$$
==> $$\displaystyle \Large pi [* (236.97 + C) - \int _{pi/2}^{3pi/2} (-1.1+6sin^2x)^2 x dx ]$$
==> $$\displaystyle \Large pi [* (236.97 + C) - \int _{pi/2}^{3pi/2} 36xsin^4x - 13.2xsin^2x + 1.21x dx ]$$
==> $$\displaystyle \Large pi [* (236.97 + C) - (36\int _{pi/2}^{3pi/2} xsin^4xdx - 13.2\int _{pi/2}^{3pi/2}xsin^2xdx + 1.21\int _{pi/2}^{3pi/2}xdx)]$$

is this right so far?

Last edited: Feb 13, 2010
2. Feb 13, 2010

### Staff: Mentor

No, it looks like you're starting off on the wrong foot here. The integral above isn't right for a couple of reasons. That first factor of x in the integral doesn't belong there, and you should be subtracting -1.1 (i.e., adding + 1.1), instead of just subtracting 1.1.
You're obviously using washers, so here's what your integral should look like.
$$\pi * \int _{\pi/2}^{3\pi/2} ((6 -(-1.1))^2 - (6sin^2x -(-1.1))^2)dx$$

In a slightly simpler form this is:
$$\pi * \int _{\pi/2}^{3\pi/2} ( 7.1^2 - (6sin^2x + 1.1)^2)dx$$
In a typical volume element ("washer"), the volume is $\pi(R^2 - r^2)\Delta x$. The larger radius R is 6 - (-1.1) = 7.1. The smaller radius r is 6sin2(x) - (-1.1) = 6sin2(x) + 1.1.

3. Feb 13, 2010

### Slimsta

i did exactly what you said and i get 451.7359646 but its wrong..
$$\displaystyle \Large pi [(50.41x + C) - (36\int _{pi/2}^{3pi/2} sin^4xdx + 13.2\int _{pi/2}^{3pi/2}sin^2xdx + 1.21x+C)]$$

== >
$$\displaystyle \Large pi [158.367 - (36\int _{pi/2}^{3pi/2} sin^4xdx + 13.2\int _{pi/2}^{3pi/2}sin^2xdx + 3.8013)]$$

then i found the antiderivative of sin4x and sin2x and got the final answer of the Volume. what did i mess up on?

4. Feb 13, 2010

### Staff: Mentor

Your answer is high by about a factor of 15. I get approximately 29.5783. After a little simplification you should have
$$\pi \int_{\pi/2}^{3\pi/2} (49.2 - 36sin^4(x) -13.2sin^2(x))dx$$

You forgot to subtract 1.21 (=1.1^2) from the 50.41 term, and it looks like you forgot that pi multiplies all of your integrals. Also, you're dealing with a definite integral: you don't need to include the constant of integration.

5. Feb 14, 2010

### Slimsta

how do you get 29.5783?
i get 256.73 and im not sure if its right or not but im pretty sure i did everything right this time.
what i forgot to do was to multiply the '-' sign inside the brackets.
$$\pi [49.2x|_{\pi/2}^{3\pi/2} - 36\int_{\pi/2}^{3\pi/2} sin^4(x)dx- 13.2\int_{\pi/2}^{3\pi/2}sin^2(x)dx]$$
$$= \pi [49.2\pi - 36(3x/2 - sin2x + sin(4x)/8)|_{\pi/2}^{3\pi/2} - 13.2/2(x - sin(2x)/2)_{\pi/2}^{3\pi/2}]$$

$$= \pi [154.57 - 48.82 - 24.03]$$ $$= \pi [81.72]$$ = 256.73

Last edited: Feb 14, 2010
6. Feb 14, 2010

### Staff: Mentor

Now I'm getting a different answer, 287.205. I'm using a computer algebra system, and I must have entered something wrong to get the lower number I posted earlier.

In your integral your terms in x are off.
$\int sin^4(x) dx = 3x/8 - 1/4*sin(2x) + 1/32*sin(4x)+C$
$\int sin^2(x) dx = x/2 - 1/4*sin(2x) + C$

Notice also that when you substitute in 3pi/2 and pi/2, all of the sine terms are zero.