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Find The Volume Of The Solid Obtained By Rotating The Region Bounded By The Given Curves About

Find the volume of the solid obtained by rotating the region bounded by the given curves?

Shell method: V = 2π ∫ r h dr {a,b} .......... limits in {}
radius is (1+y) to rotate around y=-1
= 2π ∫(1+y)(2) dx {0,1/3} + 2π ∫(1+y)(1/y - 1) dx {1/3,1}
= 4π [y + y²/2] {0,1/3} + 2π [ ln(y) - y²/2 ] {1/3,1}
= 14π/9 + 2π [ ln(3) - 4/9 ]
= 2π [ ln(3) + 1/3 ]
≈ 8.997 units³

Disc method: V = π ∫ f(x)² dx {a,b}
rotate around y=-1 extends the inner and outer radii by 1
= π ∫ (1/x + 1)² - (1)² dx {1,3} .............. note: π [(Ro)² - (Ri)²]
= π ∫ 1/x² + 2/x dx {1,3}
= π [ -1/x + 2 ln(x) ] {1,3}
= π [ -1/3 + 2 ln(3) + 1 ]
= 2π [ ln(3) + 1/3 ]
Answer: ≈ 8.997 units³

Find the volume of the solid obtained by rotating the region bounded by the curves?

1)
using the Disk Method:

about y = 1
y = x^2/36

intersection:
x^2/36 = 1
x^2 = 36
x = +/- 6

6
∫ π ( 1 - x^2/36 )^2 dx = 16π/5
0 <---- x > 0

----------------
2)
using the Disk Method:

about x-axis
y = 8x^2

1
∫ π ( 8x^2 - 0 )^2 dx = 64π/5
0 <---- when y = 0 ---> x = 0

--------------
3)
Using the shell Method:

height --------> 2 - √(x/2)
radius --------> x

intersections:
2 = √(x/2)
4 = x/2
8 = x

8
∫ 2π * x * ( 2 - √(x/2) ) dx = 128π/5
0

--------------
4)

Using the Shell Method:

4
∫ 2π * (4 - x) * ( √(x) - 0 ) = 256π/15
0

------------
5)
using the Disk Method:

about x-axis
y = x^3 ; y = x

intersection:
x^3 = x
x^3 - x = 0
x * (x^2 - 1) = 0 ----> x = 0, +/- 1

1
∫ π ( x - x^3 )^2 dx = 8π/105
0 <---- x ≥ 0

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Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified?

x = y - y^2, x = 0; about the y-axis
Washer method called for here.
Each washer will have an outer radius R = x = y - y²
and an inner radius r = 0
and a thickness of dy, so
V = ∫[0,1] π(R² - r²) dy = π∫[0,1] (y - y²)² dy = π∫[0,1] (y² - 2y³ + y^4) dy
V = π((1/3)y³ - ½y^4 + (1/5)y^5) |[0,1] = π(1/3 - 1/2 + 1/5) = (π/30)(10 - 15 + 6)
V = π/30

y = sec x, y = 1, x = -1, x = 1; about the x-axis
Here, too, washer method is called for.
Each washer will have an outer radius R = sec x
and an inner radius r = 1
and a thickness dy, so
V = ∫[a,b] π(R² - r²) = π∫[-1,1] (sec²x - 1) dx = π(tan(x) - x) |[-1,1]
Take it from here.

y = x2/3, x = 1, y = 0; about the y-axis
Shells or washers work here.
Each shell will have a radius R = x
and a height h = y = x^(2/3)
and a thickness of dx, so
V = ∫[a,b] 2πRh dx = 2π∫[0,1] x*x^(2/3) dx = 2π∫[0,1] x^(5/3) dx
V = 2π(3/8)x^(8/3) |[0,1] = 3π/4
== OR ==
Each washer will have an outer radius R = 1
and an inner radius r = x = y^(3/2)
and a thickness dy, so
V = ∫[a,b] π(R² - r²) dy = π∫[0,1] (1 - y³) dy = π(y - (1/4)y^4) |[0,1]
V = π(1 - 1/4 - 0 + 0) = 3π/4

Find the volume of the solid obtained by rotating the region bounded by the given?

First graph the situation and verify that the volume is integral (from 0 to 1) of pi(9x^2)^2 dx

Then it is pi81x^5/5 now substitute 1 and 0 and subract....

V = 81pi /5 OK!

Find the volume of the solid obtained by rotating the region bounded by the given curves about the?

USING A DISK:
dV = πy²dx
V = π⌠y²dx limit 1 -> 5
but y = √(x-1)
V = π⌠(√(x-1))²dx limit 1 -> 5
V = π⌠(x-1)dx limit 1 -> 5
V = π[x²/2-x] limit 1 -> 5
V = π[(5²/2-5) - (1²/2-1)]
V = π[(5²/2-5) - (1²/2-1)]
V = π(7.5+0.5)
V = 8π ans

USING A WASHER:
dV = 2πy(5-x)dy
V = 2π ⌠y(5-x)dy limit 0 -> 2
but y = √(x-1); x = y²+1
V = 2π ⌠y(5-(y²+1))dy limit 0 -> 2
V = 2π ⌠(4y-y³))dy limit 0 -> 2
V = 2π[4y^2/2 - y^4/4] limit 0 -> 2
V = 2π[2y^2 - y^4/4] limit 0 -> 2
V = 2π[2*2^2 - 2^4/4] - 0
V = 2π[8 - 4]
V = 8π ans

How can I find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis (y=0,y=cos(8x),x=π/16,x=0 about the axis y=−5)?

Washer MethodR=(5+cos(8x)) r=5V=Pi integral ((5+cos(8x))^2-(5)^2). Bounds 0 to pi/16

Find the volume of the solid obtained by rotating the region bounded about some line?

Since we have y=cos(6x), we should use "dx" slices and therefore the washer method (since we have a horizontal axis of revolution).
Each washer has outer radius R = cos(6x) + 7 → R² = cos²(6x) + 14cos(6x) + 49
and inner radius r = 7 → r² = 49. Then
V = ∫[a,b] π(R² - r²) dx = π ∫[0,π/12] (cos²(6x) + 14cos(6x)) dx
Using wolfram, I get
V = π(x/2 + (7/3)sin(6x) + (1/24)sin(12x) ) |[0,π/12]
V = π(π/24 + (7/3)sin(π/2) + (1/24)sinπ - (0 + (7/3)sin0 + (1/24)sin0) )
V = π(π/24 + 7/3 - 0 - (0)) = (π/24)(π + 56) ≈ 7.7 units³

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = (1/3) x^2, y = (4/3) - x^2; about the x-axis?

Find the volume [math]V[/math] of the solid obtained by rotating the region bounded by the given curves about the specified line: [math]y_1=\frac{1}{3}x^2,y_2=\frac{4}{3}-x^2;[/math] about the X axis.At the points where the curves [math]y_1[/math] and [math]y_2[/math] intersect, [math]\frac{1}{3}x^2=\frac{4}{3}-x^2.[/math][math]\Rightarrow\qquad \frac{4}{3}x^2=\frac{4}{3} \qquad\Rightarrow\qquad x^2=1 \qquad\Rightarrow\qquad x=\pm 1.[/math]In the interval [math][-1,1],y_2\ge y_1.[/math][math]\Rightarrow\qquad[/math] The required volume is the volume generated by rotating [math]y_2[/math] less the volume generated by rotating [math]y_1[/math] about the X axis.[math]\Rightarrow\qquad V=\int\limits_{-1}^1 \left(\pi y_2^2-\pi y_1^2\right)\,dx=\int\limits_{-1}^1 \pi\left(y_2^2-y_1^2\right)\,dx[/math][math]\qquad\qquad =2\pi\int\limits_{0}^1 \left(\left(\left(\frac{4}{3}-x^2\right)^2-\frac{1}{3}x^2\right)^2\right)\,dx[/math][math]\qquad\qquad =2\pi\int\limits_{0}^1 \left(\frac{16}{9}-\frac{8}{3}x^2+x^4-\frac{1}{9}x^4\right)\,dx[/math][math]\qquad\qquad =2\pi\int\limits_{0}^1 \left(\frac{16}{9}-\frac{8}{3}x^2+\frac{8}{9}x^4\right)\,dx[/math][math]\qquad\qquad =2\pi\left[\frac{16x}{9}-\frac{8}{9}x^3+\frac{8}{45}x^5\right]_{0}^1[/math][math]\qquad\qquad =2\pi\left(\frac{16}{9}-\frac{8}{9}+\frac{8}{45}\right)=\frac{96\pi}{9}.[/math]

How do I find the volume of the solid obtained by rotating the region bounded by [math]y=x^2 [/math]and [math]y=1[/math], around [math]y=7[/math]?

There are multiple ways to solve this, but I will use the "washers" method. The solid produced when rotating this region about the line [math]y=7[/math], when cut perpendicular to the x-axis, has cross sections that look like "washers", or circles with a circle cut out in the middle. To find the volume of the solid, one approach we can use is to express the area of one of these cross sections in terms of [math]x[/math], and then integrate this function from [math]x=-1 [/math]to [math]x=1[/math], since these are the endpoints of the region that we are rotating. So, essentially what we are doing is adding together the areas of the vertically oriented washers by finding the area of one washer in terms of [math]x[/math] and integrating this function from -1 to 1. This will give us the volume of the solid. Enough explaining; let's get into the math :) To find the area of a washer cross section, we will find the area of the outer circle and subtract the area of the inner circle. Since the axis of rotation is above the region we are rotating, the inner circle will be the one from the function [math]y=1[/math] and the outer circle will be the one from the function [math]y=x^2[/math] because [math]1 \geq x^2 [/math]for [math]-1 \leq x \leq 1[/math]. The radius of the inner circle is the y-value of the axis of rotation minus the y-value of the function since the cross sections are vertical. So, the radius of the inner circle is [math]7-1 = 6[/math] and the area of this circle is [math]36π[/math]. The radius of the outer circle is equal to 7 minus the y-value (in terms of x) of the function [math]y=x^2[/math] because this is the distance from the axis of rotation to the edge of the circle. Therefore, the radius is [math]7-x^2[/math] and the area of the circle is [math]π(7-x^2)^2[/math]. So, the area of a washer cross section is the area of the large circle minus the area of the small circle. Since we have already determined that we are integrating this area function from [math]x = -1[/math] to  [math]x = 1[/math], we have [math]\int_{-1}^{1} (π(7-x^2)^2 - 36π \ dx[/math]From this point, it is just a matter of solving this integral.  Here is how I did it if you are unable to solve this (I simplified it a bit at the start):It's [math]\frac{256π}{15}[/math][math] [/math] by the way ;)

How can I find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. y = sqrt (x − 1), y = 0, x = 4; about the x-axis?

It’s killing me to answer this without a blackboard to do a good explanation, but I’ll do the best I can without a good math editor on hand. You will be finding the volume using the disc method of integration. Each disc has a radius of sqrt(x-1) and a height of dx, so the volume of the solid will be the infinite sum of the volumes of all of these discs as dx becomes infinitely small, which will be the integral of the function pi sqrt(x-1) squared from 1 to 4 . Hopefully you know how to integrate because I’m skipping those details to give the answer 4.5pi.

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