The base of a solid enclosed between the graphs of y = sinx and y= -sinx from x=0 to x = pi . Each cross-section is perpendicular to the x-axis is a semicircle with the diameter connecting the two graphs . Find the volume of the solid. I know the definite integral goes from 0 to pi , a = 0, b= pi, but I'm unsure of what to do with y= sinx and y= -sinx .
the diameter of each semicircle is the distance between the two graphs:
sinx - (-sinx) = 2sinx
So, the radius of each disc is just sinx.
So, adding up the volumes of all the small semi-circular discs of thickness dx, you get
∫[0,π] 1/2 π sin^2(x) dx = π^2/4
Well, it looks like you're dealing with a pretty "sin-ful" problem! Don't worry, I'm here to help.
To find the volume of the solid, we can think of it as a bunch of little semicircles "stacked" along the x-axis. Each cross-section is a semicircle, so we just need to find the area of one of those little semicircles and then integrate it.
The diameter of each semicircle is the distance between the graphs y = sin(x) and y = -sin(x). Since both these curves are symmetrical with respect to the x-axis, we can just consider the distance from the x-axis to one of the curves, which is sin(x).
Now, the radius of each semicircle will be half of the distance between the two curves, so the radius is (1/2)sin(x).
To find the area of each semicircle, we can use the formula A = (1/2)πr^2, where r is the radius. So, the area of each individual semicircle is (1/2)π[(1/2)sin(x)]^2 = (1/8)πsin^2(x).
To find the volume of the entire solid, we integrate the area function from x = 0 to x = π:
V = ∫(0 to π) (1/8)πsin^2(x) dx.
Now, the integral of sin^2(x) can be a little tricky, but you can use a trigonometric identity to simplify it. Remember that sin^2(x) = (1/2)(1 - cos(2x)).
So, now the integral becomes:
V = ∫(0 to π) (1/8)π[(1/2)(1 - cos(2x))] dx.
You can distribute and simplify the expression inside the integral, and then integrate term by term. The final result will give you the volume of the solid.
I hope this helps! Remember, math can be a lot of fun if you look at it from the right angle (just like a funny clown!). Let me know if you have any more questions!
To find the volume of the solid, we need to integrate the area of each cross-section along the x-axis.
Since each cross-section is a semicircle, we need to find the radius of each semicircle at each x-coordinate.
The top curve is y = sinx and the bottom curve is y = -sinx. The distance between these two curves is equal to 2sinx.
The radius of each semicircle is then half of this distance, which is sinx.
Now, let's set up the integral to calculate the volume.
The cross-sectional area, A, of each semicircle is given by A = π(r^2)/2, where r is the radius.
Since r = sinx, we can write the area as A = π(sin^2x)/2.
To find the volume, we need to integrate this area over the interval from 0 to π.
Therefore, the volume of the solid can be calculated as:
V = ∫[0 to π] π(sin^2x)/2 dx
Now, we can proceed to evaluate this integral.
To find the volume of the solid, we can consider the problem as a collection of infinitely thin slices stacked on top of each other along the x-axis. Each slice is a semicircle with a diameter connecting the graphs of y = sinx and y = -sinx.
First, let's find the equation for the curve that represents the upper half of the semicircle. Since the diameter connects the graphs of y = sinx and y = -sinx, the radius of each semicircle is the difference between these two functions: r(x) = (sinx) - (-sinx) = 2sinx.
To calculate the volume of an individual slice, we need to find its area. The area of a circle is given by A = πr², but since we're dealing with semicircles, we can divide this by 2: A = (1/2)πr².
In this case, the radius of each semicircle is given by r(x) = 2sinx, so we have A(x) = (1/2)π(2sinx)² = 2πsin²x.
Now, to find the volume of the entire solid, we need to integrate the area function A(x) over the interval [0, π]. The volume V can be calculated as follows:
V = ∫[a,b] A(x) dx = ∫[0,π] 2πsin²x dx.
To evaluate this integral, we can use the power-reducing identity for sine: sin²x = (1/2)(1 - cos(2x)). Substituting this into the integral, we have:
V = ∫[0,π] 2π(1/2)(1 - cos(2x)) dx
= π∫[0,π] (1 - cos(2x)) dx.
Now, let's integrate term by term:
V = π∫[0,π] dx - π∫[0,π] cos(2x) dx.
The first integral, ∫(1) dx, is simply x, evaluated from 0 to π, which gives us x = π - 0 = π.
For the second integral, we can use the substitution u = 2x, du = 2dx:
V = π∫[0,π] dx - π∫[0,π] cos(u) (1/2) du
= π[x]∣[0,π] - (π/2)∫[0,2π] cos(u) du
= π[π - 0] - (π/2)[sin(u)]∣[0,2π]
= π² - (π/2)(sin(2π) - sin(0))
= π² - (π/2)(0 - 0)
= π².
Therefore, the volume of the solid enclosed between the graphs of y = sinx and y = -sinx from x = 0 to x = π is π².