Math / Geometry

QuestionDue Thursday $11 / 14 / 24$
1. Find the volume of a solid with the given base and cross sections. The base is a semicircley $=\sqrt{25-x^{2}}$ and the cross sections perpendicular to the $y$ -axis are triangles of equal base and height. (You must draw a picture.)

Studdy Solution

STEP 1

What is this asking? We need to find the volume of a 3D shape where the base is a half-circle and the slices, cutting across it vertically, are triangles with the same base and height. Watch out! It's easy to get confused about which way the slices are going.
Make sure you're slicing vertically, perpendicular to the $y$ -axis!
Also, remember the triangle's base and height are equal.

STEP 2

1. Draw the base
2. Figure out the triangle area
3. Integrate to find the volume

STEP 3

Alright, let's visualize this thing!
We've got a semicircle $y = \sqrt{25 - x^2}$ .
This is the top half of a circle centered at the origin with a radius of $\sqrt{25} = \textbf{5}$ . Sketch that out!

STEP 4

Now, imagine slicing this semicircle vertically, parallel to the $x$ -axis.
Each slice makes a triangle.
The base of the triangle is determined by where it hits the semicircle.
Since the semicircle is symmetrical, for any $y$ value, the $x$ value goes from $-\sqrt{25-y^2}$ to $+\sqrt{25-y^2}$ .

STEP 5

So, the base of our triangle is $2\sqrt{25-y^2}$ .
Since the triangle's base and height are equal, the height is also $2\sqrt{25-y^2}$ .

STEP 6

The area of a triangle is $\frac{1}{2} \cdot \text{base} \cdot \text{height}$ .
In our case, that's $\frac{1}{2} \cdot 2\sqrt{25-y^2} \cdot 2\sqrt{25-y^2} = 2(25-y^2)$ . So, the area of each triangular cross-section is $2(25 - y^2)$ .

STEP 7

To find the volume, we integrate the area of the cross-sections along the $y$ -axis.
Our $y$ values go from the bottom of the semicircle ( $y=0$ ) to the top ( $y=5$ ).

STEP 8

So, the volume $V$ is given by: $V = \int_{0}^{5} 2(25 - y^2) \, dy$

STEP 9

Let's evaluate this integral: $V = 2 \int_{0}^{5} (25 - y^2) \, dy = 2 \left[ 25y - \frac{1}{3}y^3 \right]_0^5$

STEP 10

Plugging in our limits of integration, we get: $V = 2 \left[ \left(25 \cdot 5 - \frac{1}{3} \cdot 5^3 \right) - \left(25 \cdot 0 - \frac{1}{3} \cdot 0^3 \right) \right]$ $V = 2 \left[ 125 - \frac{125}{3} \right] = 2 \left[ \frac{375 - 125}{3} \right] = 2 \left[ \frac{250}{3} \right] = \frac{\textbf{500}}{3}$

STEP 11

The volume of the solid is $\frac{500}{3}$ cubic units.

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QuestionDue Thursday $11 / 14 / 24$
1. Find the volume of a solid with the given base and cross sections. The base is a semicircley $=\sqrt{25-x^{2}}$ and the cross sections perpendicular to the $y$ -axis are triangles of equal base and height. (You must draw a picture.)

Studdy Solution

STEP 1

STEP 2

1. Draw the base
2. Figure out the triangle area
3. Integrate to find the volume

STEP 3

Alright, let's visualize this thing!
We've got a semicircle $y = \sqrt{25 - x^2}$ .
This is the top half of a circle centered at the origin with a radius of $\sqrt{25} = \textbf{5}$ . Sketch that out!

STEP 4

Now, imagine slicing this semicircle vertically, parallel to the $x$ -axis.
Each slice makes a triangle.
The base of the triangle is determined by where it hits the semicircle.
Since the semicircle is symmetrical, for any $y$ value, the $x$ value goes from $-\sqrt{25-y^2}$ to $+\sqrt{25-y^2}$ .

STEP 5

So, the base of our triangle is $2\sqrt{25-y^2}$ .
Since the triangle's base and height are equal, the height is also $2\sqrt{25-y^2}$ .

STEP 6

The area of a triangle is $\frac{1}{2} \cdot \text{base} \cdot \text{height}$ .
In our case, that's $\frac{1}{2} \cdot 2\sqrt{25-y^2} \cdot 2\sqrt{25-y^2} = 2(25-y^2)$ . So, the area of each triangular cross-section is $2(25 - y^2)$ .

STEP 7

To find the volume, we integrate the area of the cross-sections along the $y$ -axis.
Our $y$ values go from the bottom of the semicircle ( $y=0$ ) to the top ( $y=5$ ).

STEP 8

So, the volume $V$ is given by: $V = \int_{0}^{5} 2(25 - y^2) \, dy$

STEP 9

Let's evaluate this integral: $V = 2 \int_{0}^{5} (25 - y^2) \, dy = 2 \left[ 25y - \frac{1}{3}y^3 \right]_0^5$

STEP 10

STEP 11

The volume of the solid is $\frac{500}{3}$ cubic units.

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Studdy Solution

STEP 1

STEP 2

1. Draw the base2. Figure out the triangle area3. Integrate to find the volume

STEP 3

Alright, let's **visualize** this thing!We've got a semicircle y=25−x2y = \sqrt{25 - x^2}y=25−x2​.This is the top half of a circle centered at the origin with a radius of 25=5\sqrt{25} = \textbf{5}25​=5. **Sketch** that out!

STEP 4

STEP 5

So, the base of our triangle is 225−y22\sqrt{25-y^2}225−y2​.Since the triangle's base and height are **equal**, the height is also 225−y22\sqrt{25-y^2}225−y2​.

STEP 6

STEP 7

To find the volume, we **integrate the area** of the cross-sections along the yyy-axis.Our yyy values go from the bottom of the semicircle (y=0y=0y=0) to the top (y=5y=5y=5).

STEP 8

So, the volume VVV is given by: V=∫052(25−y2) dy V = \int_{0}^{5} 2(25 - y^2) \, dy V=∫05​2(25−y2)dy

STEP 9

Let's **evaluate this integral**: V=2∫05(25−y2) dy=2[25y−13y3]05 V = 2 \int_{0}^{5} (25 - y^2) \, dy = 2 \left[ 25y - \frac{1}{3}y^3 \right]_0^5 V=2∫05​(25−y2)dy=2[25y−31​y3]05​

STEP 10

STEP 11

The volume of the solid is 5003\frac{500}{3}3500​ cubic units.

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1. Draw the base
2. Figure out the triangle area
3. Integrate to find the volume

Alright, let's visualize this thing!
We've got a semicircle $y = \sqrt{25 - x^2}$ .
This is the top half of a circle centered at the origin with a radius of $\sqrt{25} = \textbf{5}$ . Sketch that out!

So, the base of our triangle is $2\sqrt{25-y^2}$ .
Since the triangle's base and height are equal, the height is also $2\sqrt{25-y^2}$ .

To find the volume, we integrate the area of the cross-sections along the $y$ -axis.
Our $y$ values go from the bottom of the semicircle ( $y=0$ ) to the top ( $y=5$ ).

So, the volume $V$ is given by: $V = \int_{0}^{5} 2(25 - y^2) \, dy$

Let's evaluate this integral: $V = 2 \int_{0}^{5} (25 - y^2) \, dy = 2 \left[ 25y - \frac{1}{3}y^3 \right]_0^5$

The volume of the solid is $\frac{500}{3}$ cubic units.