Trigonometric Inegrals: Volume help

[Slimsta]

Homework Statement

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.

Homework Equations

<br /> $\displaystyle \Large pi * \int _a^c x((r+yt)^2 - (r+yb)^2)dx$<br />

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?

 

[Mark44]

Homework Statement

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.

Homework Equations

<br /> $\displaystyle \Large pi * \int _a^c x((r+yt)^2 - (r+yb)^2)dx$<br />

The Attempt at a Solution

$\displaystyle \Large pi * \int _{pi/2}^{3pi/2} x((-1.1+6)^2 - (-1.1+6sin^2x)^2)dx$

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.

==> $\displaystyle \Large pi * \int _{pi/2}^{3pi/2} 24.01x - (-1.1+6sin^2x)^2 xdx$

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

is this right so far?

 

[Slimsta]

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.

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?

 

[Mark44]

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.

 

[Slimsta]

how do you get 29.5783?
i get 256.73 and I am not sure if its right or not but I am 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

 

[Mark44]

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.