Math / Calculus

QuestionUse the shell method to find the volume of the solid generated by revolving the region bounded by $x=2 y-y^{2}$ and $x=0$ about the $x$ -axis.
The volume of the solid generated by revolving the region bounded by $x=2 y-y^{2}$ and $x=0$ about the $x$ -axis is $\square$ cubic units. (Type an exact answer, using $\pi$ as needed)

Studdy Solution

STEP 1

What is this asking? We need to find the volume of the weird 3D shape we get when we spin the area between the curve $x = 2y - y^2$ and the y-axis around the x-axis. Watch out! We're spinning around the x-axis, but our function is in terms of $y$ , so things might get a little tricky!
Don't mix up the axes of rotation!

STEP 2

1. Visualize and Set Up
2. Integrate

STEP 3

First, let's sketch the curve $x = 2y - y^2$ .
Factoring gives us $x = y(2 - y)$ , so we know it's a parabola opening to the left, intersecting the y-axis at $y=0$ and $y=2$ .
The region we're interested in is between this curve and the y-axis ( $x=0$ ).

STEP 4

Since we're revolving around the x-axis, and our function is in terms of $y$ , the shell method is perfect!
Remember, the shell method formula for revolving around the x-axis is $V = 2\pi \int_a^b y \cdot f(y) \, dy$ , where $f(y)$ is the distance from the axis of rotation to the curve.
In our case, $f(y) = 2y - y^2$ .
Our limits of integration are $a=0$ and $b=2$ , the y-values where our curve intersects the y-axis.

STEP 5

Plugging everything into our shell method formula, we get: $V = 2\pi \int_0^2 y \cdot (2y - y^2) \, dy$

STEP 6

Let's simplify inside the integral: $V = 2\pi \int_0^2 (2y^2 - y^3) \, dy$

STEP 7

Now, we integrate term by term: $V = 2\pi \left[ \frac{2}{3}y^3 - \frac{1}{4}y^4 \right]_0^2$

STEP 8

Let's plug in our limits of integration: $V = 2\pi \left[ \left(\frac{2}{3}(2)^3 - \frac{1}{4}(2)^4\right) - \left(\frac{2}{3}(0)^3 - \frac{1}{4}(0)^4\right) \right]$ $V = 2\pi \left[ \left(\frac{16}{3} - \frac{16}{4}\right) - 0 \right]$ $V = 2\pi \left[ \frac{16}{3} - 4 \right]$ $V = 2\pi \left[ \frac{16}{3} - \frac{12}{3} \right]$ $V = 2\pi \left[ \frac{4}{3} \right]$

STEP 9

Finally, we multiply to get our volume: $V = \frac{8\pi}{3}$

STEP 10

The volume of the solid is $\frac{8\pi}{3}$ cubic units.

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Studdy Solution

STEP 1

STEP 2

1. Visualize and Set Up
2. Integrate

STEP 3

First, let's sketch the curve $x = 2y - y^2$ .
Factoring gives us $x = y(2 - y)$ , so we know it's a parabola opening to the left, intersecting the y-axis at $y=0$ and $y=2$ .
The region we're interested in is between this curve and the y-axis ( $x=0$ ).

STEP 4

STEP 5

Plugging everything into our shell method formula, we get: $V = 2\pi \int_0^2 y \cdot (2y - y^2) \, dy$

STEP 6

Let's simplify inside the integral: $V = 2\pi \int_0^2 (2y^2 - y^3) \, dy$

STEP 7

Now, we integrate term by term: $V = 2\pi \left[ \frac{2}{3}y^3 - \frac{1}{4}y^4 \right]_0^2$

STEP 8

STEP 9

Finally, we multiply to get our volume: $V = \frac{8\pi}{3}$

STEP 10

The volume of the solid is $\frac{8\pi}{3}$ cubic units.

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Studdy Solution

STEP 1

STEP 2

1. Visualize and Set Up2. Integrate

STEP 3

First, let's **sketch** the curve x=2y−y2x = 2y - y^2x=2y−y2.Factoring gives us x=y(2−y)x = y(2 - y)x=y(2−y), so we know it's a parabola opening to the left, intersecting the y-axis at y=0y=0y=0 and y=2y=2y=2.The region we're interested in is between this curve and the y-axis (x=0x=0x=0).

STEP 4

STEP 5

Plugging everything into our **shell method formula**, we get: V=2π∫02y⋅(2y−y2) dyV = 2\pi \int_0^2 y \cdot (2y - y^2) \, dyV=2π∫02​y⋅(2y−y2)dy

STEP 6

Let's **simplify** inside the integral: V=2π∫02(2y2−y3) dyV = 2\pi \int_0^2 (2y^2 - y^3) \, dyV=2π∫02​(2y2−y3)dy

STEP 7

Now, we **integrate** term by term: V=2π[23y3−14y4]02V = 2\pi \left[ \frac{2}{3}y^3 - \frac{1}{4}y^4 \right]_0^2V=2π[32​y3−41​y4]02​

STEP 8

STEP 9

Finally, we **multiply** to get our **volume**: V=8π3V = \frac{8\pi}{3}V=38π​

STEP 10

The volume of the solid is 8π3\frac{8\pi}{3}38π​ cubic units.

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1. Visualize and Set Up
2. Integrate

First, let's sketch the curve $x = 2y - y^2$ .
Factoring gives us $x = y(2 - y)$ , so we know it's a parabola opening to the left, intersecting the y-axis at $y=0$ and $y=2$ .
The region we're interested in is between this curve and the y-axis ( $x=0$ ).

Plugging everything into our shell method formula, we get: $V = 2\pi \int_0^2 y \cdot (2y - y^2) \, dy$

Let's simplify inside the integral: $V = 2\pi \int_0^2 (2y^2 - y^3) \, dy$

Now, we integrate term by term: $V = 2\pi \left[ \frac{2}{3}y^3 - \frac{1}{4}y^4 \right]_0^2$

Finally, we multiply to get our volume: $V = \frac{8\pi}{3}$

The volume of the solid is $\frac{8\pi}{3}$ cubic units.