✨ Magical Math ✨

Turn 2D shapes into 3D wonders with the magic of integration!
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1 The Parabolic Bowl

Find, by integration, the volume of the solid generated by revolving about the \( x \)-axis, the region enclosed by \( y = 2x^2 \), \( y = 0 \) and \( x = 1 \).
Parabolic Bowl Formation
Step 1: Identify the method - Disk Method (rotating about x-axis)
Step 2: The formula is \( V = \pi \int_{a}^{b} [f(x)]^2 dx \)
Step 3: Our function is \( f(x) = 2x^2 \), limits: \( x = 0 \) to \( x = 1 \)
Step 4: Set up the integral:
\( V = \pi \int_{0}^{1} (2x^2)^2 dx \)
Step 5: Simplify:
\( V = \pi \int_{0}^{1} 4x^4 dx \)
Step 6: Integrate:
\( V = \pi \left[ \frac{4x^5}{5} \right]_0^1 \)
Step 7: Evaluate:
\( V = \pi \left( \frac{4(1)^5}{5} - \frac{4(0)^5}{5} \right) = \frac{4\pi}{5} \)
Final Answer:
\( \boxed{\dfrac{4\pi}{5}} \)

2 The Exponential Horn

Find, by integration, the volume of the solid generated by revolving about the \( x \)-axis, the region enclosed by \( y = e^{-2x} \), \( y = 0 \), \( x = 0 \) and \( x = 1 \).
Exponential Horn
Step 1: Again, we use the Disk Method for x-axis rotation
Step 2: Formula: \( V = \pi \int_{a}^{b} [f(x)]^2 dx \)
Step 3: Our function is \( f(x) = e^{-2x} \), limits: \( x = 0 \) to \( x = 1 \)
Step 4: Set up the integral:
\( V = \pi \int_{0}^{1} (e^{-2x})^2 dx \)
Step 5: Simplify:
\( V = \pi \int_{0}^{1} e^{-4x} dx \)
Step 6: Integrate:
\( V = \pi \left[ -\frac{1}{4}e^{-4x} \right]_0^1 \)
Step 7: Evaluate:
\( V = \pi \left( -\frac{1}{4}e^{-4} - (-\frac{1}{4}e^{0}) \right) = \frac{\pi}{4}(1 - e^{-4}) \)
Final Answer:
\( \boxed{\dfrac{\pi}{4}\left(1 - e^{-4}\right)} \)

3 The Sideways Vase

Find, by integration, the volume of the solid generated by revolving about the \( y \)-axis, the region enclosed by \( x^2 = 1 + y \) and \( y = 3 \).
Sideways Vase
Step 1: This time we use the Shell Method (rotating about y-axis)
Step 2: The formula is \( V = 2\pi \int_{a}^{b} x \cdot f(x) dx \)
Step 3: Rewrite equation: \( y = x^2 - 1 \), and \( y = 3 \)
Step 4: Find intersection: \( x^2 - 1 = 3 \) → \( x = \pm 2 \)
Step 5: Height function: \( f(x) = 3 - (x^2 - 1) = 4 - x^2 \)
Step 6: Set up integral:
\( V = 2\pi \int_{0}^{2} x(4 - x^2) dx \)
(We use 0 to 2 and double due to symmetry)
Step 7: Expand:
\( V = 2\pi \int_{0}^{2} (4x - x^3) dx \)
Step 8: Integrate:
\( V = 2\pi \left[ 2x^2 - \frac{x^4}{4} \right]_0^2 \)
Step 9: Evaluate:
\( V = 2\pi \left( (8 - 4) - (0 - 0) \right) = 8\pi \)
Final Answer:
\( \boxed{8\pi} \)

4 The Donut Maker

The region enclosed between the graphs of \( y = x \) and \( y = x^2 \) is denoted by \( R \). Find the volume generated when \( R \) is rotated through \( 360^\circ \) about \( x \)-axis.
Donut-Shaped Solid
Step 1: We'll use the Washer Method (area between two curves)
Step 2: Formula: \( V = \pi \int_{a}^{b} \left( [f(x)]^2 - [g(x)]^2 \right) dx \)
Step 3: Find intersection points: \( x = x^2 \) → \( x = 0, 1 \)
Step 4: \( f(x) = x \) (upper function), \( g(x) = x^2 \) (lower function)
Step 5: Set up integral:
\( V = \pi \int_{0}^{1} \left( x^2 - (x^2)^2 \right) dx \)
Step 6: Simplify:
\( V = \pi \int_{0}^{1} (x^2 - x^4) dx \)
Step 7: Integrate:
\( V = \pi \left[ \frac{x^3}{3} - \frac{x^5}{5} \right]_0^1 \)
Step 8: Evaluate:
\( V = \pi \left( \frac{1}{3} - \frac{1}{5} \right) = \pi \left( \frac{5}{15} - \frac{3}{15} \right) = \frac{2\pi}{15} \)
Final Answer:
\( \boxed{\dfrac{2\pi}{15}} \)

5 The Conical Frustum

Find, by integration, the volume of the container which is in the shape of a right circular conical frustum.
Conical Frustum
Step 1: Imagine the frustum as a large cone minus a small cone
Step 2: Let's derive the general formula using integration
Step 3: Consider a line from (0,r1) to (h,r2) rotated about x-axis
Step 4: Equation of the line: \( y = \frac{r2-r1}{h}x + r1 \)
Step 5: Volume: \( V = \pi \int_{0}^{h} \left( \frac{r2-r1}{h}x + r1 \right)^2 dx \)
Step 6: Expand the integrand: \( \left( \frac{r2-r1}{h}x + r1 \right)^2 = \frac{(r2-r1)^2}{h^2}x^2 + \frac{2r1(r2-r1)}{h}x + r1^2 \)
Step 7: Integrate term by term:
\( V = \pi \left[ \frac{(r2-r1)^2}{3h^2}x^3 + \frac{r1(r2-r1)}{h}x^2 + r1^2x \right]_0^h \)
Step 8: Evaluate at limits:
\( V = \pi \left( \frac{(r2-r1)^2h}{3} + r1(r2-r1)h + r1^2h \right) \)
Step 9: Simplify:
\( V = \frac{\pi h}{3} (r1^2 + r1r2 + r2^2) \)
Final Answer:
\( \boxed{V = \dfrac{\pi h}{3} (r1^2 + r1r2 + r2^2)} \)

6 The Watermelon Wonder

A watermelon has an ellipsoid shape which can be obtained by revolving an ellipse with major-axis 20 cm and minor-axis 10 cm about its major-axis. Find its volume using integration.
Watermelon Ellipsoid
Step 1: Standard equation of ellipse: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
Step 2: Given: major axis = 20 cm ⇒ a = 10 cm, minor axis = 10 cm ⇒ b = 5 cm
Step 3: Solve for y: \( y = b\sqrt{1 - \frac{x^2}{a^2}} = 5\sqrt{1 - \frac{x^2}{100}} \)
Step 4: Rotating about x-axis (major axis) using Disk Method
Step 5: Volume: \( V = \pi \int_{-a}^{a} y^2 dx = \pi \int_{-10}^{10} 25\left(1 - \frac{x^2}{100}\right) dx \)
Step 6: Simplify: \( V = 25\pi \int_{-10}^{10} \left(1 - \frac{x^2}{100}\right) dx \)
Step 7: Integrate: \( V = 25\pi \left[ x - \frac{x^3}{300} \right]_{-10}^{10} \)
Step 8: Evaluate:
\( V = 25\pi \left( \left(10 - \frac{1000}{300}\right) - \left(-10 - \frac{-1000}{300}\right) \right) \)
\( V = 25\pi \left( \frac{20}{3} - (-\frac{20}{3}) \right) = 25\pi \times \frac{40}{3} = \frac{1000\pi}{3} \)
Final Answer:
\( \boxed{\dfrac{1000\pi}{3} \text{ cm}^3} \)
Magical Math Fact! ✨
The watermelon volume you just calculated is exactly 4/3πa²b for an ellipsoid rotated about its major axis. For a sphere (where a=b), this becomes the familiar 4/3πr³!