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<p>Last updated on<strong>August 5, 2025</strong></p>
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<p>Last updated on<strong>August 5, 2025</strong></p>
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<p>The surface area of revolution refers to the total area of the surface generated when a curve is revolved around a line, typically one of the coordinate axes. This concept is crucial in calculus and geometry, providing insights into the properties of 3-dimensional shapes formed by rotating a 2-dimensional curve. In this article, we will explore the surface area of revolution.</p>
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<p>The surface area of revolution refers to the total area of the surface generated when a curve is revolved around a line, typically one of the coordinate axes. This concept is crucial in calculus and geometry, providing insights into the properties of 3-dimensional shapes formed by rotating a 2-dimensional curve. In this article, we will explore the surface area of revolution.</p>
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<h2>What is the Surface Area of Revolution?</h2>
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<h2>What is the Surface Area of Revolution?</h2>
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<p>The surface area<a>of</a>revolution is the total area of the surface formed by rotating a curve around a given axis. It is measured in<a>square</a>units.</p>
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<p>The surface area<a>of</a>revolution is the total area of the surface formed by rotating a curve around a given axis. It is measured in<a>square</a>units.</p>
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<p>A common example is generating a 3D shape by rotating a 2D line or curve around the x-axis or y-axis. The resulting shape can include common geometric figures like cylinders, cones, and spheres.</p>
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<p>A common example is generating a 3D shape by rotating a 2D line or curve around the x-axis or y-axis. The resulting shape can include common geometric figures like cylinders, cones, and spheres.</p>
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<p>The surface area is calculated using integral<a>calculus</a>to account for the curve's continuous nature being revolved.</p>
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<p>The surface area is calculated using integral<a>calculus</a>to account for the curve's continuous nature being revolved.</p>
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<h2>Surface Area of Revolution Formula</h2>
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<h2>Surface Area of Revolution Formula</h2>
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<p>The<a>formula</a>for calculating the surface area of revolution depends on the axis of rotation.</p>
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<p>The<a>formula</a>for calculating the surface area of revolution depends on the axis of rotation.</p>
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<p>For a curve y = f(x) revolved around the x-axis from x = a to x = b, the formula is: Surface Area = ∫[a to b] 2πy√(1+(dy/dx)2) dx</p>
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<p>For a curve y = f(x) revolved around the x-axis from x = a to x = b, the formula is: Surface Area = ∫[a to b] 2πy√(1+(dy/dx)2) dx</p>
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<p>Similarly, for a curve x = g(y) revolved around the y-axis from y = c to y = d, the formula is: Surface Area = ∫[c to d] 2πx√(1+(dx/dy)2) dy</p>
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<p>Similarly, for a curve x = g(y) revolved around the y-axis from y = c to y = d, the formula is: Surface Area = ∫[c to d] 2πx√(1+(dx/dy)2) dy</p>
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<h2>Surface Area of Revolution about the X-axis</h2>
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<h2>Surface Area of Revolution about the X-axis</h2>
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<p>When a curve y = f(x) is revolved around the x-axis, the surface area of revolution is calculated using the formula:</p>
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<p>When a curve y = f(x) is revolved around the x-axis, the surface area of revolution is calculated using the formula:</p>
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<p>Surface Area = ∫[a to b] 2πy√(1+(dy/dx)^2) dx</p>
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<p>Surface Area = ∫[a to b] 2πy√(1+(dy/dx)^2) dx</p>
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<p>Here, y = f(x) is the<a>function</a>being revolved, and dy/dx is its derivative with respect to x.</p>
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<p>Here, y = f(x) is the<a>function</a>being revolved, and dy/dx is its derivative with respect to x.</p>
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<p>This formula accounts for the circular cross-sections formed by the rotation.</p>
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<p>This formula accounts for the circular cross-sections formed by the rotation.</p>
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<h2>Surface Area of Revolution about the Y-axis</h2>
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<h2>Surface Area of Revolution about the Y-axis</h2>
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<p>For a curve x = g(y) revolved around the y-axis, the surface area is determined by: Surface Area = ∫[c to d] 2πx√(1+(dx/dy)^2) dy</p>
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<p>For a curve x = g(y) revolved around the y-axis, the surface area is determined by: Surface Area = ∫[c to d] 2πx√(1+(dx/dy)^2) dy</p>
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<p>In this scenario, x = g(y) is the function being rotated, and dx/dy is its derivative with respect to y.</p>
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<p>In this scenario, x = g(y) is the function being rotated, and dx/dy is its derivative with respect to y.</p>
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<p>The formula considers the horizontal cross-sections made by the rotation.</p>
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<p>The formula considers the horizontal cross-sections made by the rotation.</p>
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<h2>Volume of Solid of Revolution</h2>
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<h2>Volume of Solid of Revolution</h2>
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<p>The volume of a solid of revolution is the total space occupied by the 3D shape formed when a curve is revolved around an axis. It is calculated using the disk or washer method, depending on the shape.</p>
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<p>The volume of a solid of revolution is the total space occupied by the 3D shape formed when a curve is revolved around an axis. It is calculated using the disk or washer method, depending on the shape.</p>
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<p>For example, the volume for a curve y = f(x) about the x-axis is: Volume = π∫[a to b] (f(x))^2 dx</p>
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<p>For example, the volume for a curve y = f(x) about the x-axis is: Volume = π∫[a to b] (f(x))^2 dx</p>
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<h2>Confusion between axis of rotation</h2>
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<h2>Confusion between axis of rotation</h2>
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<p>Students might confuse whether the curve is rotated about the x-axis or y-axis, leading to incorrect formula usage. Ensure clarity on the axis before applying the formula.</p>
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<p>Students might confuse whether the curve is rotated about the x-axis or y-axis, leading to incorrect formula usage. Ensure clarity on the axis before applying the formula.</p>
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<h3>Problem 1</h3>
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<h3>Problem 1</h3>
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<p>Given y = x^2, dy/dx = 2x. Surface Area = ∫[0 to 2] 2π(x^2)√(1+(2x)^2) dx = ∫[0 to 2] 2π(x^2)√(1+4x^2) dx Calculating the integral gives approximately 25.13 square units.</p>
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<p>Given y = x^2, dy/dx = 2x. Surface Area = ∫[0 to 2] 2π(x^2)√(1+(2x)^2) dx = ∫[0 to 2] 2π(x^2)√(1+4x^2) dx Calculating the integral gives approximately 25.13 square units.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Find the surface area of revolution for x = y^2 rotated about the y-axis from y = 0 to y = 1.</p>
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<p>Find the surface area of revolution for x = y^2 rotated about the y-axis from y = 0 to y = 1.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>Surface Area = 19.74 square units</p>
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<p>Surface Area = 19.74 square units</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 2</h3>
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<h3>Problem 2</h3>
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<p>Given x = y² Surface area of revolution Surface area of revolution, dx/dy = 2y. Surface Area = ∫[0 to 1] 2πy²√(1+(2y)²) dy = ∫[0 to 1] 2πy²√(1+4y²) dy Calculating the integral gives approximately 19.74 square units.</p>
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<p>Given x = y² Surface area of revolution Surface area of revolution, dx/dy = 2y. Surface Area = ∫[0 to 1] 2πy²√(1+(2y)²) dy = ∫[0 to 1] 2πy²√(1+4y²) dy Calculating the integral gives approximately 19.74 square units.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Find the surface area of revolution for y = √x rotated about the x-axis from x = 1 to x = 4.</p>
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<p>Find the surface area of revolution for y = √x rotated about the x-axis from x = 1 to x = 4.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>Surface Area = 13.68 square units</p>
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<p>Surface Area = 13.68 square units</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 3</h3>
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<h3>Problem 3</h3>
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<p>Given y = √x, dy/dx = 1/(2√x). Surface Area = ∫[1 to 4] 2π√x√(1+(1/(2√x))²) dx = ∫[1 to 4] 2π√x√(1+1/(4x)) dx Calculating the integral gives approximately 13.68 square units.</p>
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<p>Given y = √x, dy/dx = 1/(2√x). Surface Area = ∫[1 to 4] 2π√x√(1+(1/(2√x))²) dx = ∫[1 to 4] 2π√x√(1+1/(4x)) dx Calculating the integral gives approximately 13.68 square units.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Find the surface area of revolution for x = 1/y rotated about the y-axis from y = 1 to y = 3.</p>
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<p>Find the surface area of revolution for x = 1/y rotated about the y-axis from y = 1 to y = 3.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>Surface Area = 10.47 square units</p>
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<p>Surface Area = 10.47 square units</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 4</h3>
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<h3>Problem 4</h3>
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<p>Given x = 1/y, dx/dy = -1/y². Surface Area = ∫[1 to 3] 2π(1/y)√(1+(-1/y²)²) dy = ∫[1 to 3] 2π(1/y)√(1+1/y⁴) dy Calculating the integral gives approximately 10.47 square units.</p>
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<p>Given x = 1/y, dx/dy = -1/y². Surface Area = ∫[1 to 3] 2π(1/y)√(1+(-1/y²)²) dy = ∫[1 to 3] 2π(1/y)√(1+1/y⁴) dy Calculating the integral gives approximately 10.47 square units.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Find the surface area of revolution for y = 3x rotated about the x-axis from x = 0 to x = 1.</p>
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<p>Find the surface area of revolution for y = 3x rotated about the x-axis from x = 0 to x = 1.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>Surface Area = 59.22 square units</p>
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<p>Surface Area = 59.22 square units</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h2>It is the total area of the surface generated when a curve is revolved around a line, typically an axis.</h2>
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<h2>It is the total area of the surface generated when a curve is revolved around a line, typically an axis.</h2>
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<h3>1.What are the types of surface areas for revolutions?</h3>
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<h3>1.What are the types of surface areas for revolutions?</h3>
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<p>Surface areas can be calculated for revolutions about the x-axis or y-axis, depending on the curve and context.</p>
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<p>Surface areas can be calculated for revolutions about the x-axis or y-axis, depending on the curve and context.</p>
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<h3>2.What is the importance of the derivative in these formulas?</h3>
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<h3>2.What is the importance of the derivative in these formulas?</h3>
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<p>The derivative is crucial for accounting for the curve's slope, affecting the arc length element in the surface area calculation.</p>
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<p>The derivative is crucial for accounting for the curve's slope, affecting the arc length element in the surface area calculation.</p>
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<h3>3.How do you choose the correct limits of integration?</h3>
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<h3>3.How do you choose the correct limits of integration?</h3>
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<p>The limits should<a>match</a>the interval over which the curve is defined and being revolved, typically given in the problem.</p>
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<p>The limits should<a>match</a>the interval over which the curve is defined and being revolved, typically given in the problem.</p>
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<h3>4.In what units is the surface area of revolution measured?</h3>
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<h3>4.In what units is the surface area of revolution measured?</h3>
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<p>It is measured in square units like cm², m², or in².</p>
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<p>It is measured in square units like cm², m², or in².</p>
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<h2>Common Mistakes and How to Avoid Them in Surface Area of Revolution</h2>
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<h2>Common Mistakes and How to Avoid Them in Surface Area of Revolution</h2>
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<p>Students often encounter challenges in calculating the surface area of revolution due to errors in applying formulas or integrating. Below are some common mistakes and how to avoid them.</p>
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<p>Students often encounter challenges in calculating the surface area of revolution due to errors in applying formulas or integrating. Below are some common mistakes and how to avoid them.</p>
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<p>What Is Measurement? 📏 | Easy Tricks, Units & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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<p>What Is Measurement? 📏 | Easy Tricks, Units & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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<p>▶</p>
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<h2>Seyed Ali Fathima S</h2>
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<h2>Seyed Ali Fathima S</h2>
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<h3>About the Author</h3>
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<h3>About the Author</h3>
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<p>Seyed Ali Fathima S a math expert with nearly 5 years of experience as a math teacher. From an engineer to a math teacher, shows her passion for math and teaching. She is a calculator queen, who loves tables and she turns tables to puzzles and songs.</p>
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<p>Seyed Ali Fathima S a math expert with nearly 5 years of experience as a math teacher. From an engineer to a math teacher, shows her passion for math and teaching. She is a calculator queen, who loves tables and she turns tables to puzzles and songs.</p>
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<h3>Fun Fact</h3>
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<h3>Fun Fact</h3>
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<p>: She has songs for each table which helps her to remember the tables</p>
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<p>: She has songs for each table which helps her to remember the tables</p>