Find the surface area or volume.
-The region between the curve $y = \frac { 1 } { \sqrt { 1 + \cos x } } , - 1.0 \leq x \leq 0.8$, and the $x$-axis is revolved about the $x$-axis to generate a solid. Use a table of integrals to find the volume of the solid generated to two decimal places.
A) 3.20
B) 3.04
C) 3.88
D) 3.59

Question

Quizplus · Accepted Answer

To find the volume, we will use the formula of the volume of a solid of revolution:

$V = \int_a^b \pi (f(x))^2 dx$

where $a$ and $b$ are the bounds of the region to be revolved, and $f(x)$ is the function defining the curve.

In this case, we want to find the volume of the solid generated by revolving the curve $y=\frac{1}{\sqrt{1+\cos x}}$ from $x=-1$ to $x=0.8$ about the $x$-axis. So we have:

$V = \int_{-1}^{0.8} \pi \left(\frac{1}{\sqrt{1+\cos x}}ight)^2 dx$

$V = \int_{-1}^{0.8} \pi \frac{1}{1+\cos x} dx$

We can use the table of integrals to find that:

$\int \frac{1}{1+\cos x} dx = 	an\left(\frac{x}{2}ight) + C$

Plugging this back into our expression for $V$, we get:

$V = \int_{-1}^{0.8} \pi \frac{1}{1+\cos x} dx = \pi\left(	an\left(\frac{0.8}{2}ight)-	an\left(-\frac{1}{2}ight)ight) \approx 3.04$

Therefore, the answer is (B) 3.04.

Find the Surface Area or Volume y=11+cos⁡x,−1.0≤x≤0.8y = \frac { 1 } { \sqrt { 1 + \cos x } } , - 1.0 \leq x \leq 0.8y=1+cosx​1​,−1.0≤x≤0.8

Find the Surface Area or Volume $y = \frac { 1 } { \sqrt { 1 + \cos x } } , - 1.0 \leq x \leq 0.8$