Math / Calculus

Question```latex \text{Find the volume of the solid obtained by rotating the region bounded by } y = 7 \sin (x) \text{ and } y = 7 \cos (x) \text{ over the interval } 0 \leq x \leq \frac{\pi}{4} \text{ about the line } y = -1. ```

Studdy Solution

STEP 1

What is this asking? We need to find the volume of a funky shape made by spinning a curvy area around a line.
The curvy area is trapped between two wavy lines, $y = 7 \sin(x)$ and $y = 7 \cos(x)$ , from $x = 0$ to $x = \frac{\pi}{4}$ , and we're spinning it around the line $y = -1$ . Watch out! Don't forget that we're rotating around $y = -1$ , not the x-axis!
This shifts everything down, so be super careful with the radii of your washers.

STEP 2

1. Visualize and Set Up
2. Washer Method
3. Integrate
4. Evaluate

STEP 3

Alright, imagine the area between $y = 7 \sin(x)$ and $y = 7 \cos(x)$ from $x = 0$ to $x = \frac{\pi}{4}$ . $7 \cos(x)$ is on top in this region.
When we spin this around $y = -1$ , we get a cool 3D shape.

STEP 4

We'll use the washer method!
We'll slice the shape into super thin washers, find the volume of each washer, and then add up all the tiny volumes to get the total volume.

STEP 5

The outer radius, $R(x)$ , is the distance from $y = -1$ to the top curve, $y = 7 \cos(x)$ .
So, $R(x) = 7 \cos(x) - (-1) = 7 \cos(x) + 1$ .

STEP 6

The inner radius, $r(x)$ , is the distance from $y = -1$ to the bottom curve, $y = 7 \sin(x)$ .
So, $r(x) = 7 \sin(x) - (-1) = 7 \sin(x) + 1$ .

STEP 7

The area of one of these super thin washers is $\pi (R(x))^2 - \pi (r(x))^2 = \pi [(7 \cos(x) + 1)^2 - (7 \sin(x) + 1)^2]$ .

STEP 8

The volume of a single washer is its area multiplied by its tiny thickness, $dx$ .
So, the volume of a single washer is $\pi [(7 \cos(x) + 1)^2 - (7 \sin(x) + 1)^2] dx$ .

STEP 9

To find the total volume, we integrate this expression from $x = 0$ to $x = \frac{\pi}{4}$ : $V = \int_{0}^{\pi/4} \pi [(7 \cos(x) + 1)^2 - (7 \sin(x) + 1)^2] dx$

STEP 10

Let's expand the squares inside the integral: $V = \pi \int_{0}^{\pi/4} [49 \cos^2(x) + 14 \cos(x) + 1 - (49 \sin^2(x) + 14 \sin(x) + 1)] dx$

STEP 11

Simplify and combine like terms: $V = \pi \int_{0}^{\pi/4} [49(\cos^2(x) - \sin^2(x)) + 14(\cos(x) - \sin(x))] dx$

STEP 12

Remember the identity $\cos^2(x) - \sin^2(x) = \cos(2x)$ : $V = \pi \int_{0}^{\pi/4} [49 \cos(2x) + 14(\cos(x) - \sin(x))] dx$

STEP 13

Now, we can integrate: $V = \pi \left[ \frac{49}{2} \sin(2x) + 14(\sin(x) + \cos(x)) \right]_{0}^{\pi/4}$

STEP 14

Plug in the upper limit, $\frac{\pi}{4}$ : $\pi \left[ \frac{49}{2} \sin\left(\frac{\pi}{2}\right) + 14\left(\sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right)\right) \right] = \pi \left[ \frac{49}{2} + 14\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) \right] = \pi \left(\frac{49}{2} + 14\sqrt{2}\right)$

STEP 15

Plug in the lower limit, $0$ : $\pi \left[ \frac{49}{2} \sin(0) + 14(\sin(0) + \cos(0)) \right] = \pi (0 + 14(0 + 1)) = 14\pi$

STEP 16

Subtract the lower limit result from the upper limit result: $V = \pi \left(\frac{49}{2} + 14\sqrt{2}\right) - 14\pi = \pi \left(\frac{49}{2} + 14\sqrt{2} - 14\right) = \pi \left(\frac{21}{2} + 14\sqrt{2}\right)$

STEP 17

The volume of the solid is $\pi \left(\frac{21}{2} + 14\sqrt{2}\right)$ .

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Question```latex \text{Find the volume of the solid obtained by rotating the region bounded by } y = 7 \sin (x) \text{ and } y = 7 \cos (x) \text{ over the interval } 0 \leq x \leq \frac{\pi}{4} \text{ about the line } y = -1. ```

Studdy Solution

STEP 1

STEP 2

1. Visualize and Set Up
2. Washer Method
3. Integrate
4. Evaluate

STEP 3

Alright, imagine the area between $y = 7 \sin(x)$ and $y = 7 \cos(x)$ from $x = 0$ to $x = \frac{\pi}{4}$ . $7 \cos(x)$ is on top in this region.
When we spin this around $y = -1$ , we get a cool 3D shape.

STEP 4

We'll use the washer method!
We'll slice the shape into super thin washers, find the volume of each washer, and then add up all the tiny volumes to get the total volume.

STEP 5

The outer radius, $R(x)$ , is the distance from $y = -1$ to the top curve, $y = 7 \cos(x)$ .
So, $R(x) = 7 \cos(x) - (-1) = 7 \cos(x) + 1$ .

STEP 6

The inner radius, $r(x)$ , is the distance from $y = -1$ to the bottom curve, $y = 7 \sin(x)$ .
So, $r(x) = 7 \sin(x) - (-1) = 7 \sin(x) + 1$ .

STEP 7

The area of one of these super thin washers is $\pi (R(x))^2 - \pi (r(x))^2 = \pi [(7 \cos(x) + 1)^2 - (7 \sin(x) + 1)^2]$ .

STEP 8

The volume of a single washer is its area multiplied by its tiny thickness, $dx$ .
So, the volume of a single washer is $\pi [(7 \cos(x) + 1)^2 - (7 \sin(x) + 1)^2] dx$ .

STEP 9

To find the total volume, we integrate this expression from $x = 0$ to $x = \frac{\pi}{4}$ : $V = \int_{0}^{\pi/4} \pi [(7 \cos(x) + 1)^2 - (7 \sin(x) + 1)^2] dx$

STEP 10

Let's expand the squares inside the integral: $V = \pi \int_{0}^{\pi/4} [49 \cos^2(x) + 14 \cos(x) + 1 - (49 \sin^2(x) + 14 \sin(x) + 1)] dx$

STEP 11

Simplify and combine like terms: $V = \pi \int_{0}^{\pi/4} [49(\cos^2(x) - \sin^2(x)) + 14(\cos(x) - \sin(x))] dx$

STEP 12

Remember the identity $\cos^2(x) - \sin^2(x) = \cos(2x)$ : $V = \pi \int_{0}^{\pi/4} [49 \cos(2x) + 14(\cos(x) - \sin(x))] dx$

STEP 13

Now, we can integrate: $V = \pi \left[ \frac{49}{2} \sin(2x) + 14(\sin(x) + \cos(x)) \right]_{0}^{\pi/4}$

STEP 14

STEP 15

Plug in the lower limit, $0$ : $\pi \left[ \frac{49}{2} \sin(0) + 14(\sin(0) + \cos(0)) \right] = \pi (0 + 14(0 + 1)) = 14\pi$

STEP 16

Subtract the lower limit result from the upper limit result: $V = \pi \left(\frac{49}{2} + 14\sqrt{2}\right) - 14\pi = \pi \left(\frac{49}{2} + 14\sqrt{2} - 14\right) = \pi \left(\frac{21}{2} + 14\sqrt{2}\right)$

STEP 17

The volume of the solid is $\pi \left(\frac{21}{2} + 14\sqrt{2}\right)$ .

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Question```latex \text{Find the volume of the solid obtained by rotating the region bounded by } y = 7 \sin (x) \text{ and } y = 7 \cos (x) \text{ over the interval } 0 \leq x \leq \frac{\pi}{4} \text{ about the line } y = -1. ```

Studdy Solution

STEP 1

STEP 2

1. Visualize and Set Up2. Washer Method3. Integrate4. Evaluate

STEP 3

Alright, imagine the area between y=7sin⁡(x)y = 7 \sin(x)y=7sin(x) and y=7cos⁡(x)y = 7 \cos(x)y=7cos(x) from x=0x = 0x=0 to x=π4x = \frac{\pi}{4}x=4π​. 7cos⁡(x)7 \cos(x)7cos(x) is on top in this region.When we spin this around y=−1y = -1y=−1, we get a cool 3D shape.

STEP 4

We'll use the **washer method**!We'll slice the shape into super thin washers, find the volume of each washer, and then add up all the tiny volumes to get the total volume.

STEP 5

The **outer radius**, R(x)R(x)R(x), is the distance from y=−1y = -1y=−1 to the top curve, y=7cos⁡(x)y = 7 \cos(x)y=7cos(x).So, R(x)=7cos⁡(x)−(−1)=7cos⁡(x)+1R(x) = 7 \cos(x) - (-1) = 7 \cos(x) + 1R(x)=7cos(x)−(−1)=7cos(x)+1.

STEP 6

The **inner radius**, r(x)r(x)r(x), is the distance from y=−1y = -1y=−1 to the bottom curve, y=7sin⁡(x)y = 7 \sin(x)y=7sin(x).So, r(x)=7sin⁡(x)−(−1)=7sin⁡(x)+1r(x) = 7 \sin(x) - (-1) = 7 \sin(x) + 1r(x)=7sin(x)−(−1)=7sin(x)+1.

STEP 7

The **area** of one of these super thin washers is π(R(x))2−π(r(x))2=π[(7cos⁡(x)+1)2−(7sin⁡(x)+1)2]\pi (R(x))^2 - \pi (r(x))^2 = \pi [(7 \cos(x) + 1)^2 - (7 \sin(x) + 1)^2]π(R(x))2−π(r(x))2=π[(7cos(x)+1)2−(7sin(x)+1)2].

STEP 8

The **volume** of a single washer is its area multiplied by its tiny thickness, dxdxdx.So, the volume of a single washer is π[(7cos⁡(x)+1)2−(7sin⁡(x)+1)2]dx\pi [(7 \cos(x) + 1)^2 - (7 \sin(x) + 1)^2] dxπ[(7cos(x)+1)2−(7sin(x)+1)2]dx.

STEP 9

To find the **total volume**, we integrate this expression from x=0x = 0x=0 to x=π4x = \frac{\pi}{4}x=4π​: V=∫0π/4π[(7cos⁡(x)+1)2−(7sin⁡(x)+1)2]dxV = \int_{0}^{\pi/4} \pi [(7 \cos(x) + 1)^2 - (7 \sin(x) + 1)^2] dxV=∫0π/4​π[(7cos(x)+1)2−(7sin(x)+1)2]dx

STEP 10

Let's expand the squares inside the integral: V=π∫0π/4[49cos⁡2(x)+14cos⁡(x)+1−(49sin⁡2(x)+14sin⁡(x)+1)]dxV = \pi \int_{0}^{\pi/4} [49 \cos^2(x) + 14 \cos(x) + 1 - (49 \sin^2(x) + 14 \sin(x) + 1)] dxV=π∫0π/4​[49cos2(x)+14cos(x)+1−(49sin2(x)+14sin(x)+1)]dx

STEP 11

Simplify and combine like terms: V=π∫0π/4[49(cos⁡2(x)−sin⁡2(x))+14(cos⁡(x)−sin⁡(x))]dxV = \pi \int_{0}^{\pi/4} [49(\cos^2(x) - \sin^2(x)) + 14(\cos(x) - \sin(x))] dxV=π∫0π/4​[49(cos2(x)−sin2(x))+14(cos(x)−sin(x))]dx

STEP 12

Remember the identity cos⁡2(x)−sin⁡2(x)=cos⁡(2x)\cos^2(x) - \sin^2(x) = \cos(2x)cos2(x)−sin2(x)=cos(2x): V=π∫0π/4[49cos⁡(2x)+14(cos⁡(x)−sin⁡(x))]dxV = \pi \int_{0}^{\pi/4} [49 \cos(2x) + 14(\cos(x) - \sin(x))] dxV=π∫0π/4​[49cos(2x)+14(cos(x)−sin(x))]dx

STEP 13

Now, we can integrate: V=π[492sin⁡(2x)+14(sin⁡(x)+cos⁡(x))]0π/4V = \pi \left[ \frac{49}{2} \sin(2x) + 14(\sin(x) + \cos(x)) \right]_{0}^{\pi/4}V=π[249​sin(2x)+14(sin(x)+cos(x))]0π/4​

STEP 14

STEP 15

Plug in the **lower limit**, 000: π[492sin⁡(0)+14(sin⁡(0)+cos⁡(0))]=π(0+14(0+1))=14π\pi \left[ \frac{49}{2} \sin(0) + 14(\sin(0) + \cos(0)) \right] = \pi (0 + 14(0 + 1)) = 14\piπ[249​sin(0)+14(sin(0)+cos(0))]=π(0+14(0+1))=14π

STEP 16

STEP 17

The volume of the solid is π(212+142)\pi \left(\frac{21}{2} + 14\sqrt{2}\right)π(221​+142​).

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1. Visualize and Set Up
2. Washer Method
3. Integrate
4. Evaluate

Alright, imagine the area between $y = 7 \sin(x)$ and $y = 7 \cos(x)$ from $x = 0$ to $x = \frac{\pi}{4}$ . $7 \cos(x)$ is on top in this region.
When we spin this around $y = -1$ , we get a cool 3D shape.

We'll use the washer method!
We'll slice the shape into super thin washers, find the volume of each washer, and then add up all the tiny volumes to get the total volume.

The outer radius, $R(x)$ , is the distance from $y = -1$ to the top curve, $y = 7 \cos(x)$ .
So, $R(x) = 7 \cos(x) - (-1) = 7 \cos(x) + 1$ .

The inner radius, $r(x)$ , is the distance from $y = -1$ to the bottom curve, $y = 7 \sin(x)$ .
So, $r(x) = 7 \sin(x) - (-1) = 7 \sin(x) + 1$ .

The area of one of these super thin washers is $\pi (R(x))^2 - \pi (r(x))^2 = \pi [(7 \cos(x) + 1)^2 - (7 \sin(x) + 1)^2]$ .

The volume of a single washer is its area multiplied by its tiny thickness, $dx$ .
So, the volume of a single washer is $\pi [(7 \cos(x) + 1)^2 - (7 \sin(x) + 1)^2] dx$ .

To find the total volume, we integrate this expression from $x = 0$ to $x = \frac{\pi}{4}$ : $V = \int_{0}^{\pi/4} \pi [(7 \cos(x) + 1)^2 - (7 \sin(x) + 1)^2] dx$

Let's expand the squares inside the integral: $V = \pi \int_{0}^{\pi/4} [49 \cos^2(x) + 14 \cos(x) + 1 - (49 \sin^2(x) + 14 \sin(x) + 1)] dx$

Simplify and combine like terms: $V = \pi \int_{0}^{\pi/4} [49(\cos^2(x) - \sin^2(x)) + 14(\cos(x) - \sin(x))] dx$

Remember the identity $\cos^2(x) - \sin^2(x) = \cos(2x)$ : $V = \pi \int_{0}^{\pi/4} [49 \cos(2x) + 14(\cos(x) - \sin(x))] dx$

Now, we can integrate: $V = \pi \left[ \frac{49}{2} \sin(2x) + 14(\sin(x) + \cos(x)) \right]_{0}^{\pi/4}$

Plug in the lower limit, $0$ : $\pi \left[ \frac{49}{2} \sin(0) + 14(\sin(0) + \cos(0)) \right] = \pi (0 + 14(0 + 1)) = 14\pi$

The volume of the solid is $\pi \left(\frac{21}{2} + 14\sqrt{2}\right)$ .