Volumes of Solids 3
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Question 1 of 3
1. Question
Find the volume generated when `x = sqrt(16-y^2)` is rotated about the `y` – axis, between `y = -4` & `y = 4`
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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`x` `=` `sqrt(16-y^2)` `x^2` `=` `16-y^2` Square both sides Substitute `x^2` into the given formula and substitute the limits `y=-4` and `y=4``V` `=` $$\pi \int_{\color{#00880A}{-4}}^{\color{#9a00c7}{4}}$$$$x^2$$$$\:dy$$ Limits are `y=-4` and `y=4` `=` $$\pi \int_{\color{#00880A}{-4}}^{\color{#9a00c7}{4}}$$$$16-y^2$$$$\:dy$$ `x^2=16-y^2` Apply Power Rule`V` `=` $$\pi \int_{\color{#00880A}{-4}}^{\color{#9a00c7}{4}}$$$$16-y^2$$$$\:dy$$ `=` $$\pi \left(16\frac{y^{0+1}}{0+1} – \frac{y^{2+1}}{2+1} \right)$$ Apply Power Rule `=` `pi (16y-(y^3)/3)` Simplify Find the Definite Integral`V` `=` $$\pi \int_{\color{#00880A}{-4}}^{\color{#9a00c7}{4}}$$$$16-y^2$$$$\:dy$$ `=` $$\pi \left[16y – \frac{y^3}{3} \right]_{\color{#00880A}{-4}}^{\color{#9a00c7}{4}}$$ `=` $$\pi \left[\left(16(\color{#9a00c7}{4}) – \frac{\color{#9a00c7}{4}^3}{3}\right) – \left(16(\color{#00880A}{-4})-\frac{(\color{#00880A}{-4})^3}{3}\right)\right]$$ Substitute the upper `(4)` and lower limits `(-4)` `=` `pi (64-64/3)-(-64 -(-64)/3)` Simplify `=` `pi (128/3 + 128/3)` `=` `(256pi)/3` `(256pi)/3` cubic units -
Question 2 of 3
2. Question
Find the volume generated when `y = x^2` is rotated about the `y` – axis, between `y = 1` & `y = 4`
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^2` `x^2` `=` `y` Substitute `x^2` into the given formula and substitute the limits `y=1` and `y=4``V` `=` $$\pi \int_{\color{#00880A}{1}}^{\color{#9a00c7}{4}}$$$$x^2$$$$\:dy$$ Limits are `y=1` and `y=4` `=` $$\pi \int_{\color{#00880A}{1}}^{\color{#9a00c7}{4}}$$$$y$$$$\:dy$$ `x^2=y` Apply Power Rule`V` `=` $$\pi \int_{\color{#00880A}{1}}^{\color{#9a00c7}{4}}$$$$y$$$$\:dy$$ `=` $$\pi \left(\frac{y^{1+1}}{1+1}\right)$$ Apply Power Rule `=` `pi ((y^2)/2)` Simplify Find the Definite Integral`V` `=` $$\pi \int_{\color{#00880A}{1}}^{\color{#9a00c7}{4}}$$$$y$$$$\:dy$$ `=` $$\pi \left[\frac{y^2}{2} \right]_{\color{#00880A}{1}}^{\color{#9a00c7}{4}}$$ `=` $$\pi \left[\frac{\color{#9a00c7}{4}^2}{2} – \frac{\color{#00880A}{1}^2}{2}\right]$$ Substitute the upper `(4)` and lower limits `(1)` `=` `pi (16/2)-(1/2)` Simplify `=` `pi (15/2)` `=` `(15pi)/2` `(15pi)/2` cubic units -
Question 3 of 3
3. Question
Find the volume generated by the area bounded by `y = sqrtx` and `y=x^2` when rotated about the `y` – axis, between `y = 0` & `y = 1`
Hint
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Volumes-Disc Method
$$V= \pi \int_{\color{#00880A}{a}}^{\color{#9a00c7}{b}} (x_f^2 – x_c^2) dy$$First, find the values of `x_f^2` and `x_c^2`.`y` `=` `x^2` `x^2` `=` `y` `x_f^2` `=` `y` `y` `=` `sqrtx` `y^2` `=` `x` `x_c` `=` `y^2` `x_c^2` `=` `y^4` Substitute `x_f^2` and `x_c^2` into the given formula and substitute the limits `x=0` and `x=1``V` `=` $$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{1}} ($$$$x_f^2$$$$-$$$$x_c^2$$$$)\:dy$$ Limits are `x=0` and `x=1` `=` $$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{1}} ($$$$y$$$$-$$$$y^4$$$$)\:dy$$ Substitute `x_f^2` and `x_c^2` Apply Power Rule`V` `=` $$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{1}} (y-y^4)\:dy$$ `=` $$\pi \left(\frac{y^{1+1}}{1+1} – \frac{y^{4+1}}{4+1} \right)$$ Apply Power Rule `=` `pi (y^2/2 -y^5/5)` Simplify Find the Definite Integral`V` `=` $$\pi \int_{\color{#00880A}{0}}^{\color{#9a00c7}{1}} (y-y^4)\:dy$$ `=` $$\pi \left[\frac{y^2}{2} – \frac {y^5}{5} \right]_{\color{#00880A}{0}}^{\color{#9a00c7}{1}}$$ `=` $$\pi \left[\left(\frac{\color{#9a00c7}{1}^2}{2} – \frac{\color{#9a00c7}{1}^5}{5}\right) – \left(\frac{\color{#00880A}{0}^2}{2} – \frac{\color{#00880A}{0}^5}{5}\right)\right]$$ Substitute the upper `(1)` and lower limits `(0)` `=` `pi [(1/2-1/5)-(0)]` Simplify `=` `pi [3/10]` `=` `(3/10)pi` `=` `(3pi)/10` `(3pi)/10` cubic units