Mixing
Volume of a solid using disk method
$begingroup$
I have the following homework problem which I can't seem to get right. The question stated is:
Find the volume of the solid generated by revolving the region bounded
by the parabola $y=frac{x^2}{25}$ and the line y=1 about the line
y=1.
So I came up with the following integral and solved to get $frac{8}{3}pi$ but the correct answer according to my homework is $frac{16}{3}pi$.
$$piint_0^5left(1-frac{2x^2}{25}+frac{x^4}{625}right)dx$$
So is my integral correct and I'm simply not doing the math right when solving it? Or am I just missing something in the integral?
calculus integration definite-integrals volume
edited Jan 13 at 21:35
KM101
6,0251525
asked Jan 13 at 21:21
blizzblizz
1485
$endgroup$
add a comment |
$begingroup$
I have the following homework problem which I can't seem to get right. The question stated is:
Find the volume of the solid generated by revolving the region bounded
by the parabola $y=frac{x^2}{25}$ and the line y=1 about the line
y=1.
So I came up with the following integral and solved to get $frac{8}{3}pi$ but the correct answer according to my homework is $frac{16}{3}pi$.
$$piint_0^5left(1-frac{2x^2}{25}+frac{x^4}{625}right)dx$$
So is my integral correct and I'm simply not doing the math right when solving it? Or am I just missing something in the integral?
calculus integration definite-integrals volume
edited Jan 13 at 21:35
KM101
6,0251525
asked Jan 13 at 21:21
blizzblizz
1485
$endgroup$
add a comment |
$begingroup$
I have the following homework problem which I can't seem to get right. The question stated is:
Find the volume of the solid generated by revolving the region bounded
by the parabola $y=frac{x^2}{25}$ and the line y=1 about the line
y=1.
So I came up with the following integral and solved to get $frac{8}{3}pi$ but the correct answer according to my homework is $frac{16}{3}pi$.
$$piint_0^5left(1-frac{2x^2}{25}+frac{x^4}{625}right)dx$$
So is my integral correct and I'm simply not doing the math right when solving it? Or am I just missing something in the integral?
calculus integration definite-integrals volume
edited Jan 13 at 21:35
KM101
6,0251525
asked Jan 13 at 21:21
blizzblizz
1485
$endgroup$
I have the following homework problem which I can't seem to get right. The question stated is:
Find the volume of the solid generated by revolving the region bounded
by the parabola $y=frac{x^2}{25}$ and the line y=1 about the line
y=1.
So I came up with the following integral and solved to get $frac{8}{3}pi$ but the correct answer according to my homework is $frac{16}{3}pi$.
$$piint_0^5left(1-frac{2x^2}{25}+frac{x^4}{625}right)dx$$
So is my integral correct and I'm simply not doing the math right when solving it? Or am I just missing something in the integral?
calculus integration definite-integrals volume
calculus integration definite-integrals volume
edited Jan 13 at 21:35
KM101
6,0251525
asked Jan 13 at 21:21
blizzblizz
1485
edited Jan 13 at 21:35
KM101
6,0251525
asked Jan 13 at 21:21
blizzblizz
1485
edited Jan 13 at 21:35
KM101
6,0251525
edited Jan 13 at 21:35
KM101
6,0251525
edited Jan 13 at 21:35
KM101
6,0251525
6,0251525
asked Jan 13 at 21:21
blizzblizz
1485
asked Jan 13 at 21:21
blizzblizz
1485
asked Jan 13 at 21:21
blizzblizz
1485
1485
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You have to integrate from -5 to 5 or multiply your integral by two. That does the trick
answered Jan 13 at 21:41
pendermathpendermath
58612
$endgroup$
$begingroup$
But in all the other similar problems, I integrate similarly starting from zero and I still get the correct answer. And all the other similar problems also involve a revolution around both the x- and y- axes, and my answers there were all correct.
$endgroup$
– blizz
Jan 13 at 21:46
$begingroup$
What other similar problems do you have in mind? No two problems are identical. In this case, the curve crosses y=1 at points (-5,1) and (5,1). Therefore you have to integrate from -5 to 5. Does it make sense to you?
$endgroup$
– pendermath
Jan 13 at 21:48
$begingroup$
I understand...I only drew half the graph (only the first quadrant)! Thanks for clarifying!
$endgroup$
– blizz
Jan 13 at 21:54
$begingroup$
Exactly! You got it
$endgroup$
– pendermath
Jan 13 at 21:55
add a comment |
$begingroup$
You forgot to multiple it by $2$, because you set the lower bound $0$ instead of $-5$.
edited Jan 13 at 22:11
tinlyx
95221118
answered Jan 13 at 21:41
CristianoCristiano
255
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have to integrate from -5 to 5 or multiply your integral by two. That does the trick
answered Jan 13 at 21:41
pendermathpendermath
58612
$endgroup$
$begingroup$
But in all the other similar problems, I integrate similarly starting from zero and I still get the correct answer. And all the other similar problems also involve a revolution around both the x- and y- axes, and my answers there were all correct.
$endgroup$
– blizz
Jan 13 at 21:46
$begingroup$
What other similar problems do you have in mind? No two problems are identical. In this case, the curve crosses y=1 at points (-5,1) and (5,1). Therefore you have to integrate from -5 to 5. Does it make sense to you?
$endgroup$
– pendermath
Jan 13 at 21:48
$begingroup$
I understand...I only drew half the graph (only the first quadrant)! Thanks for clarifying!
$endgroup$
– blizz
Jan 13 at 21:54
$begingroup$
Exactly! You got it
$endgroup$
– pendermath
Jan 13 at 21:55
add a comment |
$begingroup$
You have to integrate from -5 to 5 or multiply your integral by two. That does the trick
answered Jan 13 at 21:41
pendermathpendermath
58612
$endgroup$
$begingroup$
But in all the other similar problems, I integrate similarly starting from zero and I still get the correct answer. And all the other similar problems also involve a revolution around both the x- and y- axes, and my answers there were all correct.
$endgroup$
– blizz
Jan 13 at 21:46
$begingroup$
What other similar problems do you have in mind? No two problems are identical. In this case, the curve crosses y=1 at points (-5,1) and (5,1). Therefore you have to integrate from -5 to 5. Does it make sense to you?
$endgroup$
– pendermath
Jan 13 at 21:48
$begingroup$
I understand...I only drew half the graph (only the first quadrant)! Thanks for clarifying!
$endgroup$
– blizz
Jan 13 at 21:54
$begingroup$
Exactly! You got it
$endgroup$
– pendermath
Jan 13 at 21:55
add a comment |
$begingroup$
You have to integrate from -5 to 5 or multiply your integral by two. That does the trick
answered Jan 13 at 21:41
pendermathpendermath
58612
$endgroup$
You have to integrate from -5 to 5 or multiply your integral by two. That does the trick
answered Jan 13 at 21:41
pendermathpendermath
58612
answered Jan 13 at 21:41
pendermathpendermath
58612
answered Jan 13 at 21:41
pendermathpendermath
58612
answered Jan 13 at 21:41
pendermathpendermath
58612
58612
$begingroup$
But in all the other similar problems, I integrate similarly starting from zero and I still get the correct answer. And all the other similar problems also involve a revolution around both the x- and y- axes, and my answers there were all correct.
$endgroup$
– blizz
Jan 13 at 21:46
$begingroup$
What other similar problems do you have in mind? No two problems are identical. In this case, the curve crosses y=1 at points (-5,1) and (5,1). Therefore you have to integrate from -5 to 5. Does it make sense to you?
$endgroup$
– pendermath
Jan 13 at 21:48
$begingroup$
I understand...I only drew half the graph (only the first quadrant)! Thanks for clarifying!
$endgroup$
– blizz
Jan 13 at 21:54
$begingroup$
Exactly! You got it
$endgroup$
– pendermath
Jan 13 at 21:55
add a comment |
$begingroup$
But in all the other similar problems, I integrate similarly starting from zero and I still get the correct answer. And all the other similar problems also involve a revolution around both the x- and y- axes, and my answers there were all correct.
$endgroup$
– blizz
Jan 13 at 21:46
$begingroup$
What other similar problems do you have in mind? No two problems are identical. In this case, the curve crosses y=1 at points (-5,1) and (5,1). Therefore you have to integrate from -5 to 5. Does it make sense to you?
$endgroup$
– pendermath
Jan 13 at 21:48
$begingroup$
I understand...I only drew half the graph (only the first quadrant)! Thanks for clarifying!
$endgroup$
– blizz
Jan 13 at 21:54
$begingroup$
Exactly! You got it
$endgroup$
– pendermath
Jan 13 at 21:55
$begingroup$
But in all the other similar problems, I integrate similarly starting from zero and I still get the correct answer. And all the other similar problems also involve a revolution around both the x- and y- axes, and my answers there were all correct.
$endgroup$
– blizz
Jan 13 at 21:46
$begingroup$
But in all the other similar problems, I integrate similarly starting from zero and I still get the correct answer. And all the other similar problems also involve a revolution around both the x- and y- axes, and my answers there were all correct.
$endgroup$
– blizz
Jan 13 at 21:46
$begingroup$
What other similar problems do you have in mind? No two problems are identical. In this case, the curve crosses y=1 at points (-5,1) and (5,1). Therefore you have to integrate from -5 to 5. Does it make sense to you?
$endgroup$
– pendermath
Jan 13 at 21:48
$begingroup$
What other similar problems do you have in mind? No two problems are identical. In this case, the curve crosses y=1 at points (-5,1) and (5,1). Therefore you have to integrate from -5 to 5. Does it make sense to you?
$endgroup$
– pendermath
Jan 13 at 21:48
$begingroup$
I understand...I only drew half the graph (only the first quadrant)! Thanks for clarifying!
$endgroup$
– blizz
Jan 13 at 21:54
$begingroup$
I understand...I only drew half the graph (only the first quadrant)! Thanks for clarifying!
$endgroup$
– blizz
Jan 13 at 21:54
$begingroup$
Exactly! You got it
$endgroup$
– pendermath
Jan 13 at 21:55
$begingroup$
Exactly! You got it
$endgroup$
– pendermath
Jan 13 at 21:55
add a comment |
$begingroup$
You forgot to multiple it by $2$, because you set the lower bound $0$ instead of $-5$.
edited Jan 13 at 22:11
tinlyx
95221118
answered Jan 13 at 21:41
CristianoCristiano
255
$endgroup$
add a comment |
$begingroup$
You forgot to multiple it by $2$, because you set the lower bound $0$ instead of $-5$.
edited Jan 13 at 22:11
tinlyx
95221118
answered Jan 13 at 21:41
CristianoCristiano
255
$endgroup$
add a comment |
$begingroup$
You forgot to multiple it by $2$, because you set the lower bound $0$ instead of $-5$.
edited Jan 13 at 22:11
tinlyx
95221118
answered Jan 13 at 21:41
CristianoCristiano
255
$endgroup$
You forgot to multiple it by $2$, because you set the lower bound $0$ instead of $-5$.
edited Jan 13 at 22:11
tinlyx
95221118
answered Jan 13 at 21:41
CristianoCristiano
255
edited Jan 13 at 22:11
tinlyx
95221118
edited Jan 13 at 22:11
tinlyx
95221118
edited Jan 13 at 22:11
tinlyx
95221118
95221118
answered Jan 13 at 21:41
CristianoCristiano
255
answered Jan 13 at 21:41
CristianoCristiano
255
answered Jan 13 at 21:41
CristianoCristiano
255
255
add a comment |
add a comment |
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