Volumes of Revolution 2
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Question 1 of 4
1. Question
Find the volume generated when `y = x^2-5x` is rotated about the `x` – axis, between `x = 0` & `x = 5`
Hint
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Volumes-Disc Method
$$V= \pi \int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} y^2 dx$$First, make `y^2` the subject of the given equation`y` `=` `x^2-5x` `y^2` `=` `(x^2-5x)^2` `y^2` `=` `x^4-10x^3+25x^2` Substitute `y^2` into the given formula and substitute the limits `x=0` and `x=5``V` `=` $$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{5}}$$$$y^2$$$$\:dx$$ Limits are `x=0` and `x=5` `=` $$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{5}}$$$$x^4-10x^3+25x^2$$$$\:dx$$ `y^2=x^4-10x^3+25x^2` Apply Power Rule`V` `=` $$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{5}}$$$$x^4-10x^3+25x^2$$$$\:dx$$ `=` $$\pi \left(\frac{x^{4+1}}{4+1} -10 \frac{x^{3+1}}{3+1} + 25 \frac {x^{2+1}}{2+1} \right)$$ Apply Power Rule `=` `pi (x^5/5 -(10x^4)/4 + (25x^3)/3)` Simplify `=` `pi (x^5/5 -(5x^4)/2 + (25x^3)/3)` Find the Definite Integral`V` `=` $$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{5}}$$$$x^4-10x^3+25x^2$$$$\:dx$$ `=` $$\pi \left[\frac{x^5}{5} – \frac {5x^4}{2} + \frac{25x^3}{3} \right]_{\color{#00880A}{0}}^{\color{#9a00c7}{5}}$$ `=` $$\pi \left[\left(\frac{\color{#9a00c7}{5}^5}{5}-\frac{5\cdot\color{#9a00c7}{5}^4}{2} + \frac{25\cdot\color{#9a00c7}{5}^3}{3}\right) – \left(\frac{\color{#00880A}{0}^5}{5}-\frac{5\cdot\color{#00880A}{0}^4}{2} + \frac{25\cdot\color{#00880A}{0}^3}{3}\right)\right]$$ Substitute the upper `(5)` and lower limits `(0)` `=` `pi [(625-(3125)/2 +(3125)/3)-(0)]` Simplify `=` `(625/6)pi` `=` `(625pi)/6` `(625pi)/6` cubic units -
Question 2 of 4
2. Question
Find the volume generated when `y = x-1` is rotated about the `y` – axis, between `y = -1` & `y = 1`
Hint
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Volumes-Disc Method
$$V= \pi \int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} x^2 dy$$First, make `x^2` the subject of the given equation`y` `=` `x-1` `y+1` `=` `x` Add `1` to both sides `x` `=` `y+1` `x^2` `=` `(y+1)^2` `x^2` `=` `y^2+2y+1` Substitute `x^2` into the given formula and substitute the limits `y=-1` and `y=1``V` `=` $$\pi \int_{\color{#00880A}{-1}}^{\color{#9a00c7}{1}}$$$$x^2$$$$\:dy$$ Limits are `y=-1` and `y=1` `=` $$\pi \int_{\color{#00880A}{-1}}^{\color{#9a00c7}{1}}$$$$y^2+2y+1$$$$\:dy$$ `x^2=y^2+2y+1` Apply Power Rule`V` `=` $$\pi \int_{\color{#00880A}{-1}}^{\color{#9a00c7}{1}}$$$$y^2+2y+1$$$$\:dy$$ `=` $$\pi \left(\frac{y^{2+1}}{2+1} +2 \frac{y^{1+1}}{1+1} + \frac {y^{0+1}}{0+1} \right)$$ Apply Power Rule `=` `pi (y^3/3 +(2y^2)/2 + y)` Simplify `=` `pi (y^3/3 +y^2 + y)` Find the Definite Integral`V` `=` $$\pi \int_{\color{#00880A}{-1}}^{\color{#9a00c7}{1}}$$$$y^2+2y+1$$$$\:dy$$ `=` $$\pi \left[\frac{y^3}{3} + y^2 + y \right]_{\color{#00880A}{-1}}^{\color{#9a00c7}{1}}$$ `=` $$\pi \left[\left(\frac{\color{#9a00c7}{1}^3}{3} +(\color{#9a00c7}{1})^2 + \color{#9a00c7}{1}\right) – \left(\frac{(\color{#00880A}{-1})^3}{3} +(\color{#00880A}{-1})^2 + \color{#00880A}{-1}\right)\right]$$ Substitute the upper `(1)` and lower limits `(-1)` `=` `pi ((1/3+2)-(-1/3))` Simplify `=` `pi (7/3-(-1/3))` `=` `(8pi)/3` `(8pi)/3` cubic units -
Question 3 of 4
3. Question
Find the volume generated when `y = x-2` is rotated about the `y` – axis, between `y = 1` & `y = 3`
Hint
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Well Done!
Incorrect
Volumes-Disc Method
$$V= \pi \int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} x^2 dy$$First, make `x^2` the subject of the given equation`y` `=` `x-2` `y+2` `=` `x` Add `2` to both sides `x` `=` `y+2` `x^2` `=` `(y+2)^2` `x^2` `=` `y^2+4y+4` Substitute `x^2` into the given formula and substitute the limits `y=1` and `y=3``V` `=` $$\pi \int_{\color{#00880A}{1}}^{\color{#9a00c7}{3}}$$$$x^2$$$$\:dy$$ Limits are `y=1` and `y=3` `=` $$\pi \int_{\color{#00880A}{1}}^{\color{#9a00c7}{3}}$$$$y^2+4y+4$$$$\:dy$$ `x^2=y^2+4y+4` Apply Power Rule`V` `=` $$\pi \int_{\color{#00880A}{1}}^{\color{#9a00c7}{3}}$$$$y^2+4y+4$$$$\:dy$$ `=` $$\pi \left(\frac{y^{2+1}}{2+1} +4 \frac{y^{1+1}}{1+1} + 4\frac {y^{0+1}}{0+1} \right)$$ Apply Power Rule `=` `pi (y^3/3 +(4y^2)/2 + 4y)` Simplify `=` `pi (y^3/3 +2y^2 + 4y)` Find the Definite Integral`V` `=` $$\pi \int_{\color{#00880A}{1}}^{\color{#9a00c7}{3}}$$$$y^2+4y+4$$$$\:dy$$ `=` $$\pi \left[\frac{y^3}{3} + 2y^2 + 4y \right]_{\color{#00880A}{1}}^{\color{#9a00c7}{3}}$$ `=` $$\pi \left[\left(\frac{\color{#9a00c7}{3}^3}{3} +2(\color{#9a00c7}{3})^2 + 4(\color{#9a00c7}{3})\right) – \left(\frac{\color{#00880A}{1}^3}{3} +2(\color{#00880A}{1})^2 + 4(\color{#00880A}{1})\right)\right]$$ Substitute the upper `(3)` and lower limits `(1)` `=` `pi ((9+18+12)-(1/3+2+4))` Simplify `=` `pi (39-(19/3))` `=` `(98pi)/3` `(98pi)/3` cubic units -
Question 4 of 4
4. Question
Find the volume generated when `y = sqrt(9-x^2)` is rotated about the `y` – axis, between `y = 0` & `y = 3`
Hint
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Excellent!
Incorrect
Volumes-Disc Method
$$V= \pi \int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} x^2 dy$$First, make `x^2` the subject of the given equation`y` `=` `sqrt(9-x^2)` `y^2` `=` `9-x^2` Square both sides `x^2` `=` `9-y^2` Solve for `x^2` Substitute `x^2` into the given formula and substitute the limits `y=0` and `y=3``V` `=` $$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{3}}$$$$x^2$$$$\:dy$$ Limits are `y=0` and `y=3` `=` $$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{3}}$$$$9-y^2$$$$\:dy$$ `x^2=9-y^2` Apply Power Rule`V` `=` $$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{3}}$$$$9-y^2$$$$\:dy$$ `=` $$\pi \left(9\frac{y^{0+1}}{0+1} – \frac{y^{2+1}}{2+1} \right)$$ Apply Power Rule `=` `pi (9y-(y^3)/3)` Simplify Find the Definite Integral`V` `=` $$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{3}}$$$$9-y^2$$$$\:dy$$ `=` $$\pi \left[9y – \frac{y^3}{3} \right]_{\color{#00880A}{0}}^{\color{#9a00c7}{3}}$$ `=` $$\pi \left[\left(9(\color{#9a00c7}{3}) – \frac{\color{#9a00c7}{3}^3}{3}\right) – \left(9(\color{#00880A}{0})-\frac{\color{#00880A}{0}^3}{3}\right)\right]$$ Substitute the upper `(3)` and lower limits `(0)` `=` `pi [(27-9)-(0)]` Simplify `=` `pi (18)` `=` `18pi` `18pi` cubic units
Quizzes
- Indefinite Integrals 1
- Indefinite Integrals 2
- Indefinite Integrals 3
- Indefinite Integrals of Exponential Functions
- Indefinite Integrals of Logarithmic Functions 1
- Indefinite Integrals of Logarithmic Functions 2
- Indefinite Integrals of Trig Functions
- Definite Integrals
- Definite Integrals of Exponential Functions
- Definite Integrals of Logarithmic Functions
- Definite Integrals of Trig Functions
- Areas Between Curves and the Axis 1
- Areas Between Curves and the Axis 2
- Area Between Curves
- Volumes of Revolution 1
- Volumes of Revolution 2
- Volumes of Revolution 3
- Trapezoidal Rule
- Simpsons Rule
- Applications of Integration for Trig Functions