Surface Area of Revolution By Integration Explained, Calculus Problems, Integral Formula, Examples

Summary

In this educational video by The Organic Chemistry Tutor, surface area of revolution is explored through various calculus examples. The video details how to integrate curves and solids by rotating them around the x-axis or y-axis. Viewers will learn techniques such as U-substitution and how to factor mathematical expressions to simplify complex calculus problems. The practical steps and detailed solutions provided aim to aid viewers in mastering the technique of integration for calculating surface areas of revolution.

Highlights

• Learn to calculate the surface area by rotating a curve around the x-axis or y-axis using integration. 🤓
• Example problems demonstrate step-by-step solutions making calculus approachable and engaging. 📏
• Explore techniques like U-substitution that simplify integration tasks in revolutionary surface area calculations. ✔️
• Understand how radius and derivatives impact calculations and problem setup in integration tasks. 🚀
• Tackle complex calculus problems with clear breakdowns and thorough explanations. 📊

Key Takeaways

• Surface area of revolution can be calculated by rotating a curve around an axis. 🤯
• Key formula involves integrating with respect to the axis of rotation, often involving U-substitution. 🔄
• Transforming and manipulating mathematical expressions is crucial in simplifying complex calculus problems. 🧮
• Understanding how to express radius in terms of x or y is essential depending on which axis you rotate around. 🔄
• Detailed breakdowns help solidify knowledge and problem-solving abilities. 📚

Overview

The Organic Chemistry Tutor is back with another insightful tutorial, this time diving into the world of calculus, specifically on how to calculate the surface area of revolution by integration. This concept is not just theoretically fascinating but is also greatly applicable in various fields involving physical sciences and engineering.

In this tutorial, viewers will follow along with example problems, enlightening us on the intricacies of handling functions, axes, and those tricky bits of calculus integration. By explaining each step and method, such as U-substitution and breaking down formulas, the tutorial offers a close-up look on how to master rotations around the x or y-axis.

Finally, the video brings clarity to one of calculus's more challenging aspects, offering learners the chance to improve their mathematical prowess and problem-solving skills. The Organic Chemistry Tutor proves once again that complex topics can be made accessible with patience and clarity, ensuring students can walk away with confidence in their newfound skills.

Chapters

• 00:00 - 02:00: Introduction to Surface Area of Revolution In this chapter, the discussion revolves around the method to calculate the surface area of a solid formed by rotating a curve around the x-axis or y-axis. The main formula introduced is the surface area, which is expressed as 2 pi times the integral from A to B of the radius (in terms of X) multiplied by the square root of 1 plus the derivative (not fully provided in the transcript).
• 02:00 - 08:00: Example 1: Rotating Curve Around X-Axis The chapter discusses the method of rotating a curve around the X-axis to find the surface area. It begins with explaining the formula to be used in calculations and provides an example with the curve y = x^3, bounded by x values from 0 to 2. The main goal is to demonstrate how to calculate the surface of the shape formed by this rotation.
• 08:00 - 13:00: Example 2: Semicircle Rotation Around X-Axis The chapter discusses the process of finding the surface area of a solid formed by the rotation of a semicircle around the x-axis. Initially, a graphical representation is considered, which focuses on the right side, where the x-values extend from zero to two. The radius, explained to be the distance between the curve and the axis of rotation, plays a crucial role in the calculation. It is noted that the radius is parallel to the direction of rotation.
• 13:00 - 20:00: Example 3: Function X in terms of Y This chapter discusses expressing the radius R in terms of X, given Y as a function of X, where Y = X^3. The explanation proceeds by stating that R(x) = X^3 in this context. It is further elaborated that if Y = X^3, then F(x) = X^3 and consequently, the derivative, F'(x), results in 3x^2. The problem then moves toward squaring both sides as a next step in the process.
• 20:00 - 25:00: Example 4: Function in Terms of X Rotated About Y-Axis In this chapter titled 'Example 4: Function in Terms of X Rotated About Y-Axis,' the focus is on finding the surface area when a function, expressed in terms of x, is rotated around the y-axis. The process involves calculating the derivative and square it, as shown with 3^2 to give 9 and x^2 to x^4. The example further develops by setting variables a and b before integrating the function between certain bounds, applying a formula based on this setup. The integration process is integral to determining the surface area.
• 25:00 - 31:00: Example 5: Cube Root Function Rotation About Y-Axis In this chapter, the focus is on rotating the Cube Root Function about the Y-Axis. The discussion begins with the transformation of the function, followed by the operation on variable S, which is multiplied by a factor involving the derivative or function prime of x squared. The expression discussed is seemingly complex, involving constants like 9 and 4, and is eventually multiplied by DX, which likely refers to an infinitesimally small change in x, usually present in calculus and integrative computations. Before proceeding further, there's a step dedicated to clearing away unnecessary elements to simplify the subsequent operations or discussions. This could involve erasing intermediate calculations, graphs, or notes that clutter the workspace or board, ensuring clarity in the demonstration that follows.
• 31:00 - 31:30: Conclusion In the conclusion chapter, the integration of the expression is discussed. The recommended technique is substitution, where 'u' is set to '1 + 9x^4'. The derivative is identified as '36x^3', suggesting a focus on simplifying the integration process using substitution.

Surface Area of Revolution By Integration Explained, Calculus Problems, Integral Formula, Examples Transcription

• 00:00 - 00:30 now in this video we're going to talk about how to calculate the surface area of a solid when rotating the curve about let's say the xaxis or even the Y AIS so here's the formula the surface area is equal to 2 pi integration from A to B times the radius in terms of X multiplied by the square root of 1 plus fime of
• 00:30 - 01:00 x 2ar * DX so that's the formula we're going to use so let's try a problem let's say that the curve is y is equal to X cub and it's bounded by the X values from 0 to two and we're going to rotate it around the X AIS calculate the surface of the
• 01:00 - 01:30 solid that is form when it's rotated about the xaxis find the surface area so let's draw a graph so we just need to focus on the right side of the graph so this is zero this is two the radius is between the curve and the axis of rotation so that's the radius it's rotating about the xais so as you can see the radius is parall parallel to the
• 01:30 - 02:00 Y AIS so r equal y now we want the radius to be in terms of X and Y is equal to X Cub so R ofx is X cub in this problem now if Y is equal to X Cub that means F ofx is X cub and frime of X is going to be 3x^2 now what we need to do is uh we need to square both sides
• 02:00 - 02:30 we need to find a square of frime of x 3^ 2 is 9 x^2 s is going to be X to 4th now let's plug in everything into the formula so we could see that a is z b is 2 so the surface area is going to be 2 pi integration from 0 to 2 R of X that's uh X cub
• 02:30 - 03:00 and then times the S < TK of 1 + frime of x^2 is 9 x 4 and then times DX so let's uh clear away a few stuff before we continue
• 03:00 - 03:30 so how can we integrate this expression what techniques can we use the best technique that we can use is use substitution let's set u = to 1 + 9x 4 du it's going to be 36 x 3r that's the derivative of 9 x
• 03:30 - 04:00 4 * DX now solving for DX it's du / 36 x 3r so now let's go ahead and replace 1 + 9x 4 with u now the square root of U we can write it as U to the half and DX we can replace that with du /
• 04:00 - 04:30 36x now we need to change the limits of integration we got to change these numbers so when X is zero let's use this equation we got to find the value of U so it's going to be 1 + 9 * U ra 4 which is just 1 plus Z so that's one now if we plug in uh two
• 04:30 - 05:00 2 to the 4th is equal to 16 and 9 * 16 that's going to be 144+ one that's 145 so now this is what we have so this is going to be one and the top number is 145 and we can cancel X Cub so let's move the 36 to the front and it's on the bottom so this is
• 05:00 - 05:30 going to be 2 piun / 36 and then we have U half du so now let's find the anti-derivative of U to the2 12 + 1 is 3 over2 so it's going to be U raised to 3 over2 and instead of dividing it by 3 over2 Let's multiply it by 2 over3 now 2 pi over 36 that reduces to pi over 18 we're going to evaluate this from
• 05:30 - 06:00 one to 145 now 18 we can break that up into 9 and two so these twos will cancel next let's multiply 9 * 3 that's 27 so what we have right now is Pi / 27 U to the 3 over 2 evaluated from 1 to
• 06:00 - 06:30 145 so that's what we currently have so now let's finish the problem let's go ahead and plug in 145 and one so this is going to be 145 to the 3es minus 1 to the 3es one ra to anything is one but now what exactly is 145 to the 3es well we can split up the exponent like this 3 over two is basically one
• 06:30 - 07:00 plus a half whenever you multiply by a common base you can add the exponents so 145 to the first power is simply 145 145 to the 1/2 power is the square otk of 145 so this is the answer now if you want the answer as a decimal that expression is approximately equal to
• 07:00 - 07:30 20344 now let's try another example so let's say that Y is equal to the < TK of 4 - x^2 and we want to find the surface area when this curve is rotated around the x-axis and it's bounded by the X values -1 and one so go ahead and try this
• 07:30 - 08:00 problem the square root of 4 - x^2 is basically a semicircle it's a semicircle with radius of two and let me draw that better that does not look like a semicircle now we only want the region starting from -1 to 1 when X is 2 Y is zero but we don't need to go that far so we're only concerned
• 08:00 - 08:30 about this area here so let's identify the radius the axis of rotation is the x-axis so the distance between the curve and the xaxis is the radius so once again we can see that the radius is equal to Y but we need it in terms of X and Y is equal to the sare < TK of 4 - x^2 so that's R
• 08:30 - 09:00 ofx now R ofx is basically the same as F ofx in this problem f ofx is 4 - x^2 to 12 now let's find the first derivative so uh frime of X is going to be 12 keep the inside the same 4 - x^2 subtract the exponent by one2 minus 1 isga 12 and according to the chain rule we need to find the derivative of the
• 09:00 - 09:30 inside the derivative of 4 - x^2 is -2X now let's simplify the expression2 and two will cancel so frime of X is equal tox / because the exponent is negative we can make it positive by moving 4 - x^2 to the bottom so this isqu < TK 4 - x^2 now we need to square both
• 09:30 - 10:00 sides we need a square of the first derivative and so fime of x^ 2 that's going to be x^2 orx * X is positive x^2 whenever you square a square root these will cancel so it's just going to be 4 - x^2 now we can write the expression for the surface area it's going to be 2 pi
• 10:00 - 10:30 integration from1 to 1 R ofx which is < TK 4 - x^2 and then times the sare < TK of 1+ the square of the derivative which is x^2 over 4 - x^2 * DX so before we continue let's just uh delete a few things
• 10:30 - 11:00 now what we're going to do is we're going to take 4 - x^2 and distribute it to these two terms because we have two square root functions we can multiply the things that are on the inside so 4 - x^2 * 1 is simply 4 - x^2 all inside the square root so we can just combine the two small square roots
• 11:00 - 11:30 into one uh big square root symbol now 4 - x^2 * x^2 over 4 - x^2 notice that the 4 - x^2 terms will cancel leaving behind just x^2 so this is going to be plus x^2 and then times DX x^2 + x^2 these will cancel they add up to zero so we have 2 pi integration from1
• 11:30 - 12:00 1 s < TK 4 DX the < TK of 4 is equal to 2 and the integration of 2dx is simply 2x 2 piun * 2 is 4 piun so we have 4 piun * X evaluated from1 to 1 so now
• 12:00 - 12:30 let's replace x with those numbers so first let's plug in the top number one and then minus the bottom number 1 1 - 1 is the same as 1 + 1 which is 2 so it's 4 Pi * 2 and so the final answer is a pi and if you want to convert that to a decimal that's going to be about 25
• 12:30 - 13:00 .33 and so that's it for this problem now let's try another one let's say that X is equal to 1/3 y^2 + 2 raised to the 3 over 2 and Y is going to be between 1 and 2 and we're going to rotate this region about the xaxis
• 13:00 - 13:30 go ahead and figure this out now it turns out that we really don't need to know exactly how the curve looks like all we need to know is the axis of rotation the curve is rotating around the x-axis let's say it looks something like this now because we have X in terms of Y we're going to integrate it using Y values that is from C to D the formula is going to be 2 pi integration from C
• 13:30 - 14:00 to D the radius is going to be in terms of Y and then times the < TK of 1 + G Prime of Y but squared so everything looks exactly the same but it's just in terms of Y instead of X but you get the same answer now the radius it's still the distance between the curve and the axis of rotation and the radius is parallel to the Y AIS which means that the radius is equal to
• 14:00 - 14:30 Y now we want the radius to be in terms of Y so we don't have to like solve for y r of Y in this problem is simply equal to y in the other examples the Rus was already y but because we wanted R ofx we needed to get it in terms of X but here it's still equal to Y but it's already in terms of Y so we're going to leave it like that now let's focus on G Prime of y g Prime of Y is equal to DX over Dy
• 14:30 - 15:00 and frime of Y I mean frime of X rather that's dy over DX which we dealt with in the last two problems let's find DX over Dy by taking the derivative of this function where x is by itself on one side of the equation so if x is equal to everything on the right side then DX over Dy is equal to 1/3 let's move this to the front so 1/3 * 3
• 15:00 - 15:30 over2 keep the inside the same and then subtract the exponent by 1 3 over 2 - 1 is 1/ 2 next according to the chain rule differentiate the inside the derivative of y^2 + 2 is 2 y so now we can cancel a 3 and we can cancel a two so DX over d Y which is
• 15:30 - 16:00 basically G Prime of y That's equal to uh y times this thing which is the same as the square < TK of y^2 + 2 now what we need to do next is we need to square both sides so y to the first Power raised to the second power that's going to be y^2 and once we square a square root this square and a square root will
• 16:00 - 16:30 cancel giving us y^2 + 2 and if we distribute y^2 it's going to be y 4 + 2 y^ 2 so that's G Prime y^ 2 so let's just write down what we have R of Y is Y and G Prime y^ 2 is y 4th + 2 y^ 2 and Y is between
• 16:30 - 17:00 two and we said one so now let's use the formula s is going to be 2 pi integration from C to D or from 1 to 2 and then it's R of Y which R of Y is just simply Y and then Square < TK 1 + G Prime of y^ 2 which is y 4th + 2 y^ 2 and then times
• 17:00 - 17:30 Dy so now let's rewrite the stuff inside the radical but in standard form so that's going to be y 4 + 2 y^ 2 + 1 * Dy notice that it's a perfect square trinomial if you have a perfect square trinomial in the form a 2 + 2 A plus b^ 2 this is equal to a + b 2 so y 4th is a 2 which means a is simply
• 17:30 - 18:00 y^2 b^2 is 1 which means B is 1 so we can Factor this trinomial and write it as y^2 + 1^ 2 so what we have is this 2 pi integration from 1 to 2 y * theare < TK of y^ 2 + 1^ 2 Dy so we can cancel this square root and a
• 18:00 - 18:30 square given us uh this expression now let's go ahead and distribute y to y^2 + 1 so that's going to be y Cub + y time Dy so the anti-derivative of Y
• 18:30 - 19:00 Cub is uh y 4th / 4 and the anti-derivative of Y the 1st power is y^2 2 evaluated from 1 to 2 and let's not forget to multiply the whole thing by 2 pi so let's plug in two so 2 to the 4th power 2 * 2 * 2 * 2 four * that's 16 and 2^ 2 is 4 now if we plug in one 1 to the 4th
• 19:00 - 19:30 power is just one so we're going to get 1 over 4 1 2 is also 1 so plus a half and then we're going to multiply everything by 2 pi so 16 over 4 is 4 and 4 / 2 is 2 then we have - 1/4 and -2 1/2 is the same as 2 over 4 now 4 + 2 is 6 -1/4 - 24 is -3
• 19:30 - 20:00 over4 now what we need to do is get common denominators 6 over 1 Let's multiply it by 4 over 4 so this is going to be 2 pi 6 * 4 is 24 and 24 - 3 is 21 now 4 / two if you divide it backwards is two so we're going to have
• 20:00 - 20:30 a two on the bottom so the answer is 21 pi/ 2 that's the final answer that is the surface area when rotating the curve about the x axis from y = 1 to Y = 2 now let's work on an example where we're going to have the function in terms of X but we're going to rotate the curve about the Y AIS so let's say that Y is equal to /4 x^2 +
• 20:30 - 21:00 1 and we want it between the X values 0 and two so if you see a range of X values then you want the radius to be in terms of X so you want everything to be in terms of X we're going to rotate this about the Y AIS x^2 opens downward and if you have a plus one it's going to be shift one unit up and we only need the right side of
• 21:00 - 21:30 the graph so this graph is going to look something like this the x intercept is two when X is 2 Y is 1 so this is the Shader region from 0 to two now because the axis of rotation is the Y AIS the radius is the distance between the y axis and
• 21:30 - 22:00 the curve so it's parallel to the x- axis but perpendicular to the Y axis because it's parallel to the x axis the radius is equal to X and we want it to be an X I mean we want it to be in terms of X so we're not going to change it now f ofx is -1/4 x^2 + 1 so now let's go ahead and find frime of x X that's going to be -4
• 22:00 - 22:30 the derivative of x^2 is 2X and the derivative of 1 is 0 so /4 * 2x is basically -2X now let's Square both sides let's find the square of the first derivative so - 12x^2 -2 * 12 that's POS 1/4 x * X is x^2 so that's what we we have now the surface
• 22:30 - 23:00 area is going to be 2 pi integration from A to B or 0 to 2 and then it's going to be R ofx which R ofx is simply x times the sare < TK of 1 + frime of x^2 which is 1/4 x^2 and then time DX so now we just need to do the math so we're going to use U
• 23:00 - 23:30 substitution let's make U = to 1 +4 X2 so du is going to be 1/4 * 2x which is 12 x * DX if we multiply both sides by 2 2D is equal to X DX and solving for DX DX is going to equal 2D / X
• 23:30 - 24:00 so the surface area is going to be 2 pi integration from 0 to 2 well we need to change uh those values using this equation so when X is zero U is equal to 1 and when X is 2 U is going to be 2^ 2 is 4 * 1/4 that's 1 + 1 U is going to be two so the new values are one and
• 24:00 - 24:30 two now let's go ahead and plug in what we have so 1 + 1/4 X2 that's U so instead of writing square root U we're going to write 12 u^ 12 and DX is uh 2D overx so let's cancel X and let's move this two let's combine it with 2 pi so this is going to be 4 pi anti-derivative 1 to
• 24:30 - 25:00 2 and then U to the half of du now the anti-derivative of U to the half is U 3 over 2 * 2/3 4 piun * 2/3 is 8 piun over 3 and if we plug in 2 and 1 into U to
• 25:00 - 25:30 the 3es it's going to be 2 to the 3 - 1 to the 32 2 to the 32 is basically 2 to the 1 power * 2 to half so that's 2 > 2 now we can write the answer like this or we could distribute the eight if we want to 8 * 2 is 16 and so it's going to be 16 < tk2 minus 8 so you can leave the answer like that now the decimal value of this answer is
• 25:30 - 26:00 about 15. 318 so this is the solution so this is going to be the last example for this uh video let's say that Y is equal to the cube root of x and the Y value ranges from 1 to two let's rotate the region about the Y AIS and let's calculate the surface area so the graph of cubot of x just the right portion looks like this on the
• 26:00 - 26:30 left side it looks like that but we only need the graph from one to two on the Y AIS so let's say this is one and this is two now we're going to rotate the curve about the y- axis so therefore this is r r is the distance between the axis of rotation and a curve so because R is parallel to the X
• 26:30 - 27:00 AIS R is equal to X but notice that the range consists of Y values which indicates that we want R not in terms of X but in terms of Y so somehow we have to use this equation to get R in terms of Y so Y is equal to the cube root of x so let's raise both sides to the third power we need to solve for x so these will cancel and we can see that
• 27:00 - 27:30 X is equal to YB so if the radius is equal to X then the radius is also equal to Y Cub now let's focus on G Prime of Y so using this equation if x is equal to Y Cub then DX over Dy is equal to 3 y^2 which is the same as G Prime of Y now we need to find
• 27:30 - 28:00 G Prime of y^ 2 so that's uh 3 y^ 2^ S which is uh 9 y 4 so now let's write the equation for the surface area let's set up the integral it's going to be 2 pi integration from C to D which C is 1 D is 2 so from 1 to two and then it's R of Y which is y cub time < TK 1 + G Prime y^ 2 which is 9 y
• 28:00 - 28:30 4times Dy so once we set up the integral now we just got to worry about the math part so let's use U substitution let's make U = to 1 + 9 y 4 so du is going to be 36 YB Dy solving for d Dy it's going to be equal to this
• 28:30 - 29:00 thing du over 36 YB now we need to change the X values into u values using this equation so I mean when Y is one I said X values but I meant y values when Y is one U is going to be basically 10 and when Y is 2 2 4 is 16 16 * 9 is
• 29:00 - 29:30 144 + 1 that's 145 so the surface area is going to be 2 pi 10 to 145 y^ the 3 U to the half and then Dy is this it's du over 36 y Cub so let's cancel these two let's take this move it to the front 2 Pi /
• 29:30 - 30:00 36 is basically pi over 18 the anti-derivative of U to the half is U 3 over2 * 2/3 now we know that 2 over 18 is going to be 1 over9 and 1 over 9 * piun over 3 is pi over 27 and then we're going to evaluate this from 10 to
• 30:00 - 30:30 145 so this is going to be pi over 27 and then 145 rais to the 3 2 - 10^ 3 2 which we know it's going to be 145 < TK 145 - 10 < TK 10 so this is the exact answer the decimal equivalent is going to be it's approximately 1
• 30:30 - 31:00 99.48 and so that's it for this video thanks for watching