Question

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line.

y = 2 sin(x), y = 2 cos(x), 0≤x≤pi/4; about y = -1

Answer 1

The volume V of the solid obtained by rotating the region bounded by the given curves about the specified line is
`& V=4 \pi\left(√(2)-(1)/(2)\right)`.

We can apply the washer approach to find the volume of the solid formed by rotating the region bordered by two curves about an external horizontal axis.

The volume (V) formula for the washer method is as follows:

V =
`\pi \int_a^b\left[(f(x))^2-(g(x))^2\right] d x`

where:

- f(x) represents the outer radius.

- g(x) represents the inner radius.

The distances between the axis of rotation and the outer and inner curves, respectively, are known as the outer and inner radiuses.

In this instance, on the interval
`$0 \leq x \leq \pi / 4$`, the outer curve is
`$y=2 \cos (x)$`, while the inner curve is
`$y=2 \sin (x)$`. Hence, given that are revolving about y=-1,

f(x)=2 cos (x)+1

g(x)=2 sin (x)+1

So, the integral for the volume using the washer method is:

`$$V=\pi \int_0^{(\pi)/(4)}\left[(2 \cos (x)+1)^2-(2 \sin (x)+1)^2\right] d x$$`

Do, simplification of
`$(2 \cos (x)+1)^2-(2 \sin (x)+1)^2$` :

`(2 \cos (x)+1)^2-(2 \sin (x)+1)^2 \\= & \left(4 \cos ^2(x)+4 \cos (x)+1\right)-\left(4 \sin ^2(x)+4 \sin (x)+1\right) \\= & 4 \cos ^2(x)+4 \cos (x)-4 \sin ^2(x)-4 \sin (x) \\= & 4\left(\cos ^2(x)-\sin ^2(x)\right)+4 \cos (x)-4 \sin (x) \\= & 4 \cos (2 x)+4 \cos (x)-4 \sin (x)\end{aligned}$$`

Hence,

`V=\pi \int_0^{(\pi)/(4)}[4 \cos (2 x)+4 \cos (x)-4 \sin (x)] d x`

After solving:

`$$V=4 \pi \int_0^{(\pi)/(4)}[\cos (2 x)+\cos (x)-\sin (x)] d x$$`

Now, Analyse the integral.

`& V=4 \pi\left[(1)/(2) \sin (2 x)+\sin (x)+\cos (x)\right]_0^{(\pi)/(4)} \\& V=4 \pi\left[\left((1)/(2) \sin \left(2 \cdot (\pi)/(4)\right)+\sin \left((\pi)/(4)\right)+\cos \left((\pi)/(4)\right)\right)-\left((1)/(2) \sin (2 \cdot 0)+\sin (0)+\cos (0)\right)\right] \\`

`& V=4 \pi\left[\left((1)/(2) \sin \left((\pi)/(2)\right)+\sin \left((\pi)/(4)\right)+\cos \left((\pi)/(4)\right)\right)-\left((1)/(2) \sin (0)+\sin (0)+\cos (0)\right)\right] \\& V=4 \pi\left[\left((1)/(2)+(√(2))/(2)+(√(2))/(2)\right)-(0+0+1)\right] \\`

`& V=4 \pi\left[(1)/(2)+√(2)-1\right] \\& V=4 \pi\left(√(2)-(1)/(2)\right)`

Answer 2

`\displaystyle V = 4\pi \left(√(2)-(1)/(2)\right)\approx 11.488\; \sf (3\;d.p.)`

Explanation:

To find the volume of the solid obtained by rotating the region bounded by two curves about an external horizontal axis, we can use the washer method.

The washer method formula for volume (V) is given by:

`\displaystyle V = \pi \int_a^b \left[ (f(x))^2 - (g(x))^2 \right] \;dx`

where:

• f(x) represents the outer radius.
• g(x) represents the inner radius.

The outer radius is the distance from the axis of rotation to the outer curve, and the inner radius is the distance from the axis of rotation to the inner curve.

In this case, the outer curve is y = 2cos(x) and the inner curve is y = 2sin(x) on the interval 0 ≤ x ≤ π/4. So, as are rotating about y = -1, then:

`f(x) = 2\cos(x) + 1`

`g(x) = 2\sin(x) + 1`

Therefore, the integral for the volume using the washer method is:

`\displaystyle V = \pi \int_0^{(\pi)/(4)} \left[ (2\cos(x) + 1)^2 - (2\sin(x) + 1)^2 \right] \;dx`

First, simplify (2cos(x) + 1)² - (2sin(x) + 1)²:

`\begin{aligned}&\;\;\;\;\:(2\cos(x) + 1)^2 - (2\sin(x) + 1)^2\\\\&=(4\cos^2(x)+4\cos(x)+1)-(4\sin^2(x)+4\sin(x)+1)\\\\&=4\cos^2(x)+4\cos(x)-4\sin^2(x)-4\sin(x)\\\\&=4(\cos^2(x)-\sin^2(x))+4\cos(x)-4\sin(x)\\\\&=4\cos(2x)+4\cos(x)-4\sin(x)\end{aligned}`

Therefore:

`\displaystyle V = \pi \int_0^{(\pi)/(4)} \left[ 4\cos(2x)+4\cos(x)-4\sin(x) \right] \;dx`

Take out the constant:

`\displaystyle V = 4\pi \int_0^{(\pi)/(4)} \left[\cos(2x)+\cos(x)-\sin(x) \right] \;dx`

Evaluate the integral:

`\displaystyle V = 4\pi \left[ (1)/(2)\sin(2x)+\sin(x)+\cos(x) \right]^{(\pi)/(4)}_0`

`\displaystyle V = 4\pi \left[ \left((1)/(2)\sin\left(2\cdot (\pi)/(4)\right)+\sin\left((\pi)/(4)\right)+\cos\left((\pi)/(4)\right)\right)-\left((1)/(2)\sin(2\cdot 0)+\sin(0)+\cos(0) \right)\right]`

`\displaystyle V = 4\pi \left[ \left((1)/(2)\sin\left((\pi)/(2)\right)+\sin\left((\pi)/(4)\right)+\cos\left((\pi)/(4)\right)\right)-\left((1)/(2)\sin(0)+\sin(0)+\cos(0) \right)\right]`

`\displaystyle V = 4\pi \left[ \left((1)/(2)+(√(2))/(2)+(√(2))/(2)\right)-\left(0+0+1 \right)\right]`

`\displaystyle V = 4\pi \left[(1)/(2)+√(2)-1\right]`

`\displaystyle V = 4\pi \left(√(2)-(1)/(2)\right)`

Therefore, the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line is:

`\displaystyle V = 4\pi \left(√(2)-(1)/(2)\right)\approx 11.488\; \sf (3\;d.p.)`