Volumes of Revolution 2
Try VividMath Premium to unlock full access
Time limit: 0
Quiz summary
0 of 4 questions completed
Questions:
- 1
- 2
- 3
- 4
Information
–
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
- 1
- 2
- 3
- 4
- Answered
- Review
-
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
Help VideoCorrect
Keep Going!
Incorrect
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
Help VideoCorrect
Great Work!
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-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
Help VideoCorrect
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
Help VideoCorrect
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
- Definite Integrals
- Areas Between Curves and the Axis 1
- Areas Between Curves and the Axis 2
- The Area Between Curves
- Volumes of Revolution 1
- Volumes of Revolution 2
- Volumes of Revolution 3
- Trapezoidal Rule
- Simpsons Rule
- Integral of a Trigonometric Function 1
- Integral of a Trigonometric Function 2
- Applications of Integration for Trig Functions