Mixing
Why doesn't the second method work?
$begingroup$
I have been trying to complete this question:
The region R, bounded by the curve with equation $y=sin(x)$, $0leqslant x leqslant pi$ and the line with equation $y = dfrac{1}{sqrt2}$
The region R is rotated through $2pi$ radians about the line $y = dfrac{1}{sqrt2}$
Show that the solid of revolution formed has area $dfrac{pi}{2}(pi-3)$
I have tried two methods, one of which has worked, and the other has not, but I don't understand why the second method didn't work
The first method was to do the integral $$piint_{dfrac{pi}{4}}^{dfrac{3pi}{4}} (sin^2x-dfrac{1}{sqrt2})^2 dx$$, which worked fine.
The second method was to do the integral $$piint_{dfrac{pi}{4}}^{dfrac{3pi}{4}} sin^2x dx - piint_{dfrac{pi}{4}}^{dfrac{3pi}{4}} dfrac{1}{2} dx$$, which would find the volume from rotating $y=sin x$, and subtract the volume from rotating $y=dfrac{1}{sqrt2}$, which I thought would find the correct value, but it didn't.
I can see that the integrals are different, because the expansion of $(sin^2x-dfrac{1}{sqrt2})^2$ $ne$ $sin^2x-dfrac{1}{sqrt2}$, but I am not getting why the second method doesn't work, as it looks like it should, graphically.
My question is why doesn't the second method work, but the first does?
calculus integration solid-of-revolution
asked Jan 26 at 16:42
Joshua PeachamJoshua Peacham
162
$endgroup$
add a comment |
$begingroup$
I have been trying to complete this question:
The region R, bounded by the curve with equation $y=sin(x)$, $0leqslant x leqslant pi$ and the line with equation $y = dfrac{1}{sqrt2}$
The region R is rotated through $2pi$ radians about the line $y = dfrac{1}{sqrt2}$
Show that the solid of revolution formed has area $dfrac{pi}{2}(pi-3)$
I have tried two methods, one of which has worked, and the other has not, but I don't understand why the second method didn't work
The first method was to do the integral $$piint_{dfrac{pi}{4}}^{dfrac{3pi}{4}} (sin^2x-dfrac{1}{sqrt2})^2 dx$$, which worked fine.
The second method was to do the integral $$piint_{dfrac{pi}{4}}^{dfrac{3pi}{4}} sin^2x dx - piint_{dfrac{pi}{4}}^{dfrac{3pi}{4}} dfrac{1}{2} dx$$, which would find the volume from rotating $y=sin x$, and subtract the volume from rotating $y=dfrac{1}{sqrt2}$, which I thought would find the correct value, but it didn't.
I can see that the integrals are different, because the expansion of $(sin^2x-dfrac{1}{sqrt2})^2$ $ne$ $sin^2x-dfrac{1}{sqrt2}$, but I am not getting why the second method doesn't work, as it looks like it should, graphically.
My question is why doesn't the second method work, but the first does?
calculus integration solid-of-revolution
asked Jan 26 at 16:42
Joshua PeachamJoshua Peacham
162
$endgroup$
$begingroup$
Surely what you're calculating is a volume, not area.
$endgroup$
– J.G.
Jan 26 at 17:09
1
$begingroup$
@J.G. i meant volumes, sorry
$endgroup$
– Joshua Peacham
Jan 26 at 17:15
add a comment |
$begingroup$
I have been trying to complete this question:
The region R, bounded by the curve with equation $y=sin(x)$, $0leqslant x leqslant pi$ and the line with equation $y = dfrac{1}{sqrt2}$
The region R is rotated through $2pi$ radians about the line $y = dfrac{1}{sqrt2}$
Show that the solid of revolution formed has area $dfrac{pi}{2}(pi-3)$
I have tried two methods, one of which has worked, and the other has not, but I don't understand why the second method didn't work
The first method was to do the integral $$piint_{dfrac{pi}{4}}^{dfrac{3pi}{4}} (sin^2x-dfrac{1}{sqrt2})^2 dx$$, which worked fine.
The second method was to do the integral $$piint_{dfrac{pi}{4}}^{dfrac{3pi}{4}} sin^2x dx - piint_{dfrac{pi}{4}}^{dfrac{3pi}{4}} dfrac{1}{2} dx$$, which would find the volume from rotating $y=sin x$, and subtract the volume from rotating $y=dfrac{1}{sqrt2}$, which I thought would find the correct value, but it didn't.
I can see that the integrals are different, because the expansion of $(sin^2x-dfrac{1}{sqrt2})^2$ $ne$ $sin^2x-dfrac{1}{sqrt2}$, but I am not getting why the second method doesn't work, as it looks like it should, graphically.
My question is why doesn't the second method work, but the first does?
calculus integration solid-of-revolution
asked Jan 26 at 16:42
Joshua PeachamJoshua Peacham
162
$endgroup$
I have been trying to complete this question:
The region R, bounded by the curve with equation $y=sin(x)$, $0leqslant x leqslant pi$ and the line with equation $y = dfrac{1}{sqrt2}$
The region R is rotated through $2pi$ radians about the line $y = dfrac{1}{sqrt2}$
Show that the solid of revolution formed has area $dfrac{pi}{2}(pi-3)$
I have tried two methods, one of which has worked, and the other has not, but I don't understand why the second method didn't work
The first method was to do the integral $$piint_{dfrac{pi}{4}}^{dfrac{3pi}{4}} (sin^2x-dfrac{1}{sqrt2})^2 dx$$, which worked fine.
The second method was to do the integral $$piint_{dfrac{pi}{4}}^{dfrac{3pi}{4}} sin^2x dx - piint_{dfrac{pi}{4}}^{dfrac{3pi}{4}} dfrac{1}{2} dx$$, which would find the volume from rotating $y=sin x$, and subtract the volume from rotating $y=dfrac{1}{sqrt2}$, which I thought would find the correct value, but it didn't.
I can see that the integrals are different, because the expansion of $(sin^2x-dfrac{1}{sqrt2})^2$ $ne$ $sin^2x-dfrac{1}{sqrt2}$, but I am not getting why the second method doesn't work, as it looks like it should, graphically.
My question is why doesn't the second method work, but the first does?
calculus integration solid-of-revolution
calculus integration solid-of-revolution
asked Jan 26 at 16:42
Joshua PeachamJoshua Peacham
162
asked Jan 26 at 16:42
Joshua PeachamJoshua Peacham
162
edited Jan 26 at 17:16
Joshua Peacham
asked Jan 26 at 16:42
Joshua PeachamJoshua Peacham
162
asked Jan 26 at 16:42
Joshua PeachamJoshua Peacham
162
asked Jan 26 at 16:42
Joshua PeachamJoshua Peacham
162
162
$begingroup$
Surely what you're calculating is a volume, not area.
$endgroup$
– J.G.
Jan 26 at 17:09
1
$begingroup$
@J.G. i meant volumes, sorry
$endgroup$
– Joshua Peacham
Jan 26 at 17:15
add a comment |
$begingroup$
Surely what you're calculating is a volume, not area.
$endgroup$
– J.G.
Jan 26 at 17:09
1
$begingroup$
@J.G. i meant volumes, sorry
$endgroup$
– Joshua Peacham
Jan 26 at 17:15
$begingroup$
Surely what you're calculating is a volume, not area.
$endgroup$
– J.G.
Jan 26 at 17:09
$begingroup$
Surely what you're calculating is a volume, not area.
$endgroup$
– J.G.
Jan 26 at 17:09
1
1
$begingroup$
@J.G. i meant volumes, sorry
$endgroup$
– Joshua Peacham
Jan 26 at 17:15
$begingroup$
@J.G. i meant volumes, sorry
$endgroup$
– Joshua Peacham
Jan 26 at 17:15
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Where did you find this question? The second integral seems to me the correct way to evaluate the volume.
When calculating the volum of a revolution solid using this method, what are you doing in fact is "adding" the areas of infinite cilinders with very small heights, each one with radius $f(x)$ and "height" $dx$. The "volume" of each cilinder is $pi[f(x)]²dx$.
So the volume of the solid is
$$V=piint_a^b [f(x)]² dx$$.
(You can make this rigorous using Riemann sums, but I think it's unnecessary here.)
In your case, you're not adding cylinders, you are adding "cylinders" with a hole, so the "volume" of each cylinder is $[pi f(x)²-pi g(x)²]dx$. "Adding it up", it would result in
$$V=piint_a^b [f(x)]² dx-piint_a^b [g(x)]² dx$$
answered Jan 26 at 17:08
Célio AugustoCélio Augusto
193
$endgroup$
$begingroup$
It's from my school textbook, I thought the second method should work as it finds the total volume from rotating y=sinx then subtracting the cylinder that we don't want, as it is the part outside the region R
$endgroup$
– Joshua Peacham
Jan 26 at 17:14
add a comment |
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$begingroup$
Where did you find this question? The second integral seems to me the correct way to evaluate the volume.
When calculating the volum of a revolution solid using this method, what are you doing in fact is "adding" the areas of infinite cilinders with very small heights, each one with radius $f(x)$ and "height" $dx$. The "volume" of each cilinder is $pi[f(x)]²dx$.
So the volume of the solid is
$$V=piint_a^b [f(x)]² dx$$.
(You can make this rigorous using Riemann sums, but I think it's unnecessary here.)
In your case, you're not adding cylinders, you are adding "cylinders" with a hole, so the "volume" of each cylinder is $[pi f(x)²-pi g(x)²]dx$. "Adding it up", it would result in
$$V=piint_a^b [f(x)]² dx-piint_a^b [g(x)]² dx$$
answered Jan 26 at 17:08
Célio AugustoCélio Augusto
193
$endgroup$
$begingroup$
It's from my school textbook, I thought the second method should work as it finds the total volume from rotating y=sinx then subtracting the cylinder that we don't want, as it is the part outside the region R
$endgroup$
– Joshua Peacham
Jan 26 at 17:14
add a comment |
$begingroup$
Where did you find this question? The second integral seems to me the correct way to evaluate the volume.
When calculating the volum of a revolution solid using this method, what are you doing in fact is "adding" the areas of infinite cilinders with very small heights, each one with radius $f(x)$ and "height" $dx$. The "volume" of each cilinder is $pi[f(x)]²dx$.
So the volume of the solid is
$$V=piint_a^b [f(x)]² dx$$.
(You can make this rigorous using Riemann sums, but I think it's unnecessary here.)
In your case, you're not adding cylinders, you are adding "cylinders" with a hole, so the "volume" of each cylinder is $[pi f(x)²-pi g(x)²]dx$. "Adding it up", it would result in
$$V=piint_a^b [f(x)]² dx-piint_a^b [g(x)]² dx$$
answered Jan 26 at 17:08
Célio AugustoCélio Augusto
193
$endgroup$
$begingroup$
It's from my school textbook, I thought the second method should work as it finds the total volume from rotating y=sinx then subtracting the cylinder that we don't want, as it is the part outside the region R
$endgroup$
– Joshua Peacham
Jan 26 at 17:14
add a comment |
$begingroup$
Where did you find this question? The second integral seems to me the correct way to evaluate the volume.
When calculating the volum of a revolution solid using this method, what are you doing in fact is "adding" the areas of infinite cilinders with very small heights, each one with radius $f(x)$ and "height" $dx$. The "volume" of each cilinder is $pi[f(x)]²dx$.
So the volume of the solid is
$$V=piint_a^b [f(x)]² dx$$.
(You can make this rigorous using Riemann sums, but I think it's unnecessary here.)
In your case, you're not adding cylinders, you are adding "cylinders" with a hole, so the "volume" of each cylinder is $[pi f(x)²-pi g(x)²]dx$. "Adding it up", it would result in
$$V=piint_a^b [f(x)]² dx-piint_a^b [g(x)]² dx$$
answered Jan 26 at 17:08
Célio AugustoCélio Augusto
193
$endgroup$
Where did you find this question? The second integral seems to me the correct way to evaluate the volume.
When calculating the volum of a revolution solid using this method, what are you doing in fact is "adding" the areas of infinite cilinders with very small heights, each one with radius $f(x)$ and "height" $dx$. The "volume" of each cilinder is $pi[f(x)]²dx$.
So the volume of the solid is
$$V=piint_a^b [f(x)]² dx$$.
(You can make this rigorous using Riemann sums, but I think it's unnecessary here.)
In your case, you're not adding cylinders, you are adding "cylinders" with a hole, so the "volume" of each cylinder is $[pi f(x)²-pi g(x)²]dx$. "Adding it up", it would result in
$$V=piint_a^b [f(x)]² dx-piint_a^b [g(x)]² dx$$
answered Jan 26 at 17:08
Célio AugustoCélio Augusto
193
edited Jan 26 at 17:15
answered Jan 26 at 17:08
Célio AugustoCélio Augusto
193
answered Jan 26 at 17:08
Célio AugustoCélio Augusto
193
answered Jan 26 at 17:08
Célio AugustoCélio Augusto
193
193
$begingroup$
It's from my school textbook, I thought the second method should work as it finds the total volume from rotating y=sinx then subtracting the cylinder that we don't want, as it is the part outside the region R
$endgroup$
– Joshua Peacham
Jan 26 at 17:14
add a comment |
$begingroup$
It's from my school textbook, I thought the second method should work as it finds the total volume from rotating y=sinx then subtracting the cylinder that we don't want, as it is the part outside the region R
$endgroup$
– Joshua Peacham
Jan 26 at 17:14
$begingroup$
It's from my school textbook, I thought the second method should work as it finds the total volume from rotating y=sinx then subtracting the cylinder that we don't want, as it is the part outside the region R
$endgroup$
– Joshua Peacham
Jan 26 at 17:14
$begingroup$
It's from my school textbook, I thought the second method should work as it finds the total volume from rotating y=sinx then subtracting the cylinder that we don't want, as it is the part outside the region R
$endgroup$
– Joshua Peacham
Jan 26 at 17:14
add a comment |
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$begingroup$
Surely what you're calculating is a volume, not area.
$endgroup$
– J.G.
Jan 26 at 17:09
1
$begingroup$
@J.G. i meant volumes, sorry
$endgroup$
– Joshua Peacham
Jan 26 at 17:15