Mixing
Simplest Way to Evaluate Lengthy Integration Equation with Succinct Answer
$begingroup$
Recently, I determined the volume of the solid of revolution created by rotating the parabola, $y=ax^2$, about the axis of revolution, $y=bx$, as a function of $a$ and $b$, where both variables are positive, real numbers to be
$$V=frac{pi b^5}{30a^3sqrt{1+b^2}}tag{1}$$
I proved this several different ways, but I got stuck at the conclusion of my application of the disk method. The final, integrable equation that I derived was
$$V=int_0^{frac{b}{a}sqrt{1+b^2}}pileft(-frac{1}{b}x-frac{sqrt{1+b^2}}{2ab^2}+frac{sqrt{1+b^2}}{2ab^2}sqrt{4abxsqrt{1+b^2}+1}right)^2dxtag{2}$$
I happen to know that equation 2 does simplify to equation 1 (thank you, Desmos). However, I could not think of a way of performing this simplification without a full-blown expansion of the integrand, separation at the plus/minus signs, integration, returning of limits, and simplification, which (I began but did not finish) took a head-exploding amount of care.
In effect, my question is: What is the most conservative, simplest, best way of evaluating equation 2? All solutions are welcome, but one that uses relatively elementary calculus concepts would be best (I'm a junior in high school taking AP BC Calculus).
Thanks!
calculus algebra-precalculus volume
asked Jan 1 at 20:00
Shady PuckShady Puck
1285
$endgroup$
add a comment |
$begingroup$
Recently, I determined the volume of the solid of revolution created by rotating the parabola, $y=ax^2$, about the axis of revolution, $y=bx$, as a function of $a$ and $b$, where both variables are positive, real numbers to be
$$V=frac{pi b^5}{30a^3sqrt{1+b^2}}tag{1}$$
I proved this several different ways, but I got stuck at the conclusion of my application of the disk method. The final, integrable equation that I derived was
$$V=int_0^{frac{b}{a}sqrt{1+b^2}}pileft(-frac{1}{b}x-frac{sqrt{1+b^2}}{2ab^2}+frac{sqrt{1+b^2}}{2ab^2}sqrt{4abxsqrt{1+b^2}+1}right)^2dxtag{2}$$
I happen to know that equation 2 does simplify to equation 1 (thank you, Desmos). However, I could not think of a way of performing this simplification without a full-blown expansion of the integrand, separation at the plus/minus signs, integration, returning of limits, and simplification, which (I began but did not finish) took a head-exploding amount of care.
In effect, my question is: What is the most conservative, simplest, best way of evaluating equation 2? All solutions are welcome, but one that uses relatively elementary calculus concepts would be best (I'm a junior in high school taking AP BC Calculus).
Thanks!
calculus algebra-precalculus volume
asked Jan 1 at 20:00
Shady PuckShady Puck
1285
$endgroup$
$begingroup$
In your link to "proved this several ways", $b$ was $1.$
$endgroup$
– coffeemath
Jan 1 at 20:22
1
$begingroup$
@coffeemath Yes — I just meant that I applied the cone, shell, and disk methods described in that question and its answers. I apologize for the confusion.
$endgroup$
– Shady Puck
Jan 1 at 20:46
$begingroup$
Shady Puck-- Just finished getting the rotation volume using Pappus' Theorem [find centroid, get distance $d$ from there to rotation axis, then it's $2pi d A$ with $A$ the area of rotated region]. I got exactly your formula (1). I'm curious how you set up your disc method-- looks like an involved expression to integrate; I assume discs are perpendicular to rotation axis?
$endgroup$
– coffeemath
Jan 2 at 19:45
add a comment |
$begingroup$
Recently, I determined the volume of the solid of revolution created by rotating the parabola, $y=ax^2$, about the axis of revolution, $y=bx$, as a function of $a$ and $b$, where both variables are positive, real numbers to be
$$V=frac{pi b^5}{30a^3sqrt{1+b^2}}tag{1}$$
I proved this several different ways, but I got stuck at the conclusion of my application of the disk method. The final, integrable equation that I derived was
$$V=int_0^{frac{b}{a}sqrt{1+b^2}}pileft(-frac{1}{b}x-frac{sqrt{1+b^2}}{2ab^2}+frac{sqrt{1+b^2}}{2ab^2}sqrt{4abxsqrt{1+b^2}+1}right)^2dxtag{2}$$
I happen to know that equation 2 does simplify to equation 1 (thank you, Desmos). However, I could not think of a way of performing this simplification without a full-blown expansion of the integrand, separation at the plus/minus signs, integration, returning of limits, and simplification, which (I began but did not finish) took a head-exploding amount of care.
In effect, my question is: What is the most conservative, simplest, best way of evaluating equation 2? All solutions are welcome, but one that uses relatively elementary calculus concepts would be best (I'm a junior in high school taking AP BC Calculus).
Thanks!
calculus algebra-precalculus volume
asked Jan 1 at 20:00
Shady PuckShady Puck
1285
$endgroup$
Recently, I determined the volume of the solid of revolution created by rotating the parabola, $y=ax^2$, about the axis of revolution, $y=bx$, as a function of $a$ and $b$, where both variables are positive, real numbers to be
$$V=frac{pi b^5}{30a^3sqrt{1+b^2}}tag{1}$$
I proved this several different ways, but I got stuck at the conclusion of my application of the disk method. The final, integrable equation that I derived was
$$V=int_0^{frac{b}{a}sqrt{1+b^2}}pileft(-frac{1}{b}x-frac{sqrt{1+b^2}}{2ab^2}+frac{sqrt{1+b^2}}{2ab^2}sqrt{4abxsqrt{1+b^2}+1}right)^2dxtag{2}$$
I happen to know that equation 2 does simplify to equation 1 (thank you, Desmos). However, I could not think of a way of performing this simplification without a full-blown expansion of the integrand, separation at the plus/minus signs, integration, returning of limits, and simplification, which (I began but did not finish) took a head-exploding amount of care.
In effect, my question is: What is the most conservative, simplest, best way of evaluating equation 2? All solutions are welcome, but one that uses relatively elementary calculus concepts would be best (I'm a junior in high school taking AP BC Calculus).
Thanks!
calculus algebra-precalculus volume
calculus algebra-precalculus volume
asked Jan 1 at 20:00
Shady PuckShady Puck
1285
asked Jan 1 at 20:00
Shady PuckShady Puck
1285
asked Jan 1 at 20:00
Shady PuckShady Puck
1285
asked Jan 1 at 20:00
Shady PuckShady Puck
1285
asked Jan 1 at 20:00
Shady PuckShady Puck
1285
1285
$begingroup$
In your link to "proved this several ways", $b$ was $1.$
$endgroup$
– coffeemath
Jan 1 at 20:22
1
$begingroup$
@coffeemath Yes — I just meant that I applied the cone, shell, and disk methods described in that question and its answers. I apologize for the confusion.
$endgroup$
– Shady Puck
Jan 1 at 20:46
$begingroup$
Shady Puck-- Just finished getting the rotation volume using Pappus' Theorem [find centroid, get distance $d$ from there to rotation axis, then it's $2pi d A$ with $A$ the area of rotated region]. I got exactly your formula (1). I'm curious how you set up your disc method-- looks like an involved expression to integrate; I assume discs are perpendicular to rotation axis?
$endgroup$
– coffeemath
Jan 2 at 19:45
add a comment |
$begingroup$
In your link to "proved this several ways", $b$ was $1.$
$endgroup$
– coffeemath
Jan 1 at 20:22
1
$begingroup$
@coffeemath Yes — I just meant that I applied the cone, shell, and disk methods described in that question and its answers. I apologize for the confusion.
$endgroup$
– Shady Puck
Jan 1 at 20:46
$begingroup$
Shady Puck-- Just finished getting the rotation volume using Pappus' Theorem [find centroid, get distance $d$ from there to rotation axis, then it's $2pi d A$ with $A$ the area of rotated region]. I got exactly your formula (1). I'm curious how you set up your disc method-- looks like an involved expression to integrate; I assume discs are perpendicular to rotation axis?
$endgroup$
– coffeemath
Jan 2 at 19:45
$begingroup$
In your link to "proved this several ways", $b$ was $1.$
$endgroup$
– coffeemath
Jan 1 at 20:22
$begingroup$
In your link to "proved this several ways", $b$ was $1.$
$endgroup$
– coffeemath
Jan 1 at 20:22
1
1
$begingroup$
@coffeemath Yes — I just meant that I applied the cone, shell, and disk methods described in that question and its answers. I apologize for the confusion.
$endgroup$
– Shady Puck
Jan 1 at 20:46
$begingroup$
@coffeemath Yes — I just meant that I applied the cone, shell, and disk methods described in that question and its answers. I apologize for the confusion.
$endgroup$
– Shady Puck
Jan 1 at 20:46
$begingroup$
Shady Puck-- Just finished getting the rotation volume using Pappus' Theorem [find centroid, get distance $d$ from there to rotation axis, then it's $2pi d A$ with $A$ the area of rotated region]. I got exactly your formula (1). I'm curious how you set up your disc method-- looks like an involved expression to integrate; I assume discs are perpendicular to rotation axis?
$endgroup$
– coffeemath
Jan 2 at 19:45
$begingroup$
Shady Puck-- Just finished getting the rotation volume using Pappus' Theorem [find centroid, get distance $d$ from there to rotation axis, then it's $2pi d A$ with $A$ the area of rotated region]. I got exactly your formula (1). I'm curious how you set up your disc method-- looks like an involved expression to integrate; I assume discs are perpendicular to rotation axis?
$endgroup$
– coffeemath
Jan 2 at 19:45
add a comment |
1 Answer
1
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$begingroup$
Use computer algebra (e.g., Mathematica) which gives a very simple answer:
$$frac{pi b^5}{30 sqrt{b^2+1}}$$
answered Jan 1 at 20:06
David G. StorkDavid G. Stork
10.2k21332
$endgroup$
$begingroup$
Well, yes, that would be very simple. I was asking for a alternate step-by-step solution that did not require the care of a full expansion. BTW, you may want to check your input (as your solution does not match my equation 1).
$endgroup$
– Shady Puck
Jan 1 at 20:11
$begingroup$
@ShadyPuck: I re-checked my code indeed implements your Eq. (2). Perhaps you should re-check it.
$endgroup$
– David G. Stork
Jan 1 at 20:14
add a comment |
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1 Answer
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$begingroup$
Use computer algebra (e.g., Mathematica) which gives a very simple answer:
$$frac{pi b^5}{30 sqrt{b^2+1}}$$
answered Jan 1 at 20:06
David G. StorkDavid G. Stork
10.2k21332
$endgroup$
$begingroup$
Well, yes, that would be very simple. I was asking for a alternate step-by-step solution that did not require the care of a full expansion. BTW, you may want to check your input (as your solution does not match my equation 1).
$endgroup$
– Shady Puck
Jan 1 at 20:11
$begingroup$
@ShadyPuck: I re-checked my code indeed implements your Eq. (2). Perhaps you should re-check it.
$endgroup$
– David G. Stork
Jan 1 at 20:14
add a comment |
$begingroup$
Use computer algebra (e.g., Mathematica) which gives a very simple answer:
$$frac{pi b^5}{30 sqrt{b^2+1}}$$
answered Jan 1 at 20:06
David G. StorkDavid G. Stork
10.2k21332
$endgroup$
$begingroup$
Well, yes, that would be very simple. I was asking for a alternate step-by-step solution that did not require the care of a full expansion. BTW, you may want to check your input (as your solution does not match my equation 1).
$endgroup$
– Shady Puck
Jan 1 at 20:11
$begingroup$
@ShadyPuck: I re-checked my code indeed implements your Eq. (2). Perhaps you should re-check it.
$endgroup$
– David G. Stork
Jan 1 at 20:14
add a comment |
$begingroup$
Use computer algebra (e.g., Mathematica) which gives a very simple answer:
$$frac{pi b^5}{30 sqrt{b^2+1}}$$
answered Jan 1 at 20:06
David G. StorkDavid G. Stork
10.2k21332
$endgroup$
Use computer algebra (e.g., Mathematica) which gives a very simple answer:
$$frac{pi b^5}{30 sqrt{b^2+1}}$$
answered Jan 1 at 20:06
David G. StorkDavid G. Stork
10.2k21332
answered Jan 1 at 20:06
David G. StorkDavid G. Stork
10.2k21332
answered Jan 1 at 20:06
David G. StorkDavid G. Stork
10.2k21332
answered Jan 1 at 20:06
David G. StorkDavid G. Stork
10.2k21332
10.2k21332
$begingroup$
Well, yes, that would be very simple. I was asking for a alternate step-by-step solution that did not require the care of a full expansion. BTW, you may want to check your input (as your solution does not match my equation 1).
$endgroup$
– Shady Puck
Jan 1 at 20:11
$begingroup$
@ShadyPuck: I re-checked my code indeed implements your Eq. (2). Perhaps you should re-check it.
$endgroup$
– David G. Stork
Jan 1 at 20:14
add a comment |
$begingroup$
Well, yes, that would be very simple. I was asking for a alternate step-by-step solution that did not require the care of a full expansion. BTW, you may want to check your input (as your solution does not match my equation 1).
$endgroup$
– Shady Puck
Jan 1 at 20:11
$begingroup$
@ShadyPuck: I re-checked my code indeed implements your Eq. (2). Perhaps you should re-check it.
$endgroup$
– David G. Stork
Jan 1 at 20:14
$begingroup$
Well, yes, that would be very simple. I was asking for a alternate step-by-step solution that did not require the care of a full expansion. BTW, you may want to check your input (as your solution does not match my equation 1).
$endgroup$
– Shady Puck
Jan 1 at 20:11
$begingroup$
Well, yes, that would be very simple. I was asking for a alternate step-by-step solution that did not require the care of a full expansion. BTW, you may want to check your input (as your solution does not match my equation 1).
$endgroup$
– Shady Puck
Jan 1 at 20:11
$begingroup$
@ShadyPuck: I re-checked my code indeed implements your Eq. (2). Perhaps you should re-check it.
$endgroup$
– David G. Stork
Jan 1 at 20:14
$begingroup$
@ShadyPuck: I re-checked my code indeed implements your Eq. (2). Perhaps you should re-check it.
$endgroup$
– David G. Stork
Jan 1 at 20:14
add a comment |
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$begingroup$
In your link to "proved this several ways", $b$ was $1.$
$endgroup$
– coffeemath
Jan 1 at 20:22
1
$begingroup$
@coffeemath Yes — I just meant that I applied the cone, shell, and disk methods described in that question and its answers. I apologize for the confusion.
$endgroup$
– Shady Puck
Jan 1 at 20:46
$begingroup$
Shady Puck-- Just finished getting the rotation volume using Pappus' Theorem [find centroid, get distance $d$ from there to rotation axis, then it's $2pi d A$ with $A$ the area of rotated region]. I got exactly your formula (1). I'm curious how you set up your disc method-- looks like an involved expression to integrate; I assume discs are perpendicular to rotation axis?
$endgroup$
– coffeemath
Jan 2 at 19:45