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Evaluation of $int (r^2-x^2) dx$
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I am learning how to approximate the volume of a sphere with integrals. I am confused on how $(1/3)x^3$ became $(1/3)r^3$, and why this being an even function allows you to pull a $2$ out front. Please help, many thanks
enter image description here
derivatives definite-integrals
edited Jul 20 at 21:45
amWhy
189k25219431
asked Jul 20 at 21:09
C_bri
112
add a comment |Â
up vote
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I am learning how to approximate the volume of a sphere with integrals. I am confused on how $(1/3)x^3$ became $(1/3)r^3$, and why this being an even function allows you to pull a $2$ out front. Please help, many thanks
enter image description here
derivatives definite-integrals
edited Jul 20 at 21:45
amWhy
189k25219431
asked Jul 20 at 21:09
C_bri
112
Substitute $r$ instead of $x$.
– Nosrati
Jul 20 at 21:14
At first upper bound $colorredminus$ the second lower bound!
– Nosrati
Jul 20 at 21:17
$$left(r^2x-dfrac13x^3right)_0^r=left(r^2times r-dfrac13(r)^3right)-left(r^2times0-dfrac13(0)^3right)$$ final answer is $dfrac43pi r^3$.
– Nosrati
Jul 20 at 21:26
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am learning how to approximate the volume of a sphere with integrals. I am confused on how $(1/3)x^3$ became $(1/3)r^3$, and why this being an even function allows you to pull a $2$ out front. Please help, many thanks
enter image description here
derivatives definite-integrals
edited Jul 20 at 21:45
amWhy
189k25219431
asked Jul 20 at 21:09
C_bri
112
I am learning how to approximate the volume of a sphere with integrals. I am confused on how $(1/3)x^3$ became $(1/3)r^3$, and why this being an even function allows you to pull a $2$ out front. Please help, many thanks
enter image description here
derivatives definite-integrals
edited Jul 20 at 21:45
amWhy
189k25219431
asked Jul 20 at 21:09
C_bri
112
edited Jul 20 at 21:45
amWhy
189k25219431
edited Jul 20 at 21:45
amWhy
189k25219431
edited Jul 20 at 21:45
amWhy
189k25219431
189k25219431
asked Jul 20 at 21:09
C_bri
112
asked Jul 20 at 21:09
C_bri
112
asked Jul 20 at 21:09
C_bri
112
112
Substitute $r$ instead of $x$.
– Nosrati
Jul 20 at 21:14
At first upper bound $colorredminus$ the second lower bound!
– Nosrati
Jul 20 at 21:17
$$left(r^2x-dfrac13x^3right)_0^r=left(r^2times r-dfrac13(r)^3right)-left(r^2times0-dfrac13(0)^3right)$$ final answer is $dfrac43pi r^3$.
– Nosrati
Jul 20 at 21:26
add a comment |Â
Substitute $r$ instead of $x$.
– Nosrati
Jul 20 at 21:14
At first upper bound $colorredminus$ the second lower bound!
– Nosrati
Jul 20 at 21:17
$$left(r^2x-dfrac13x^3right)_0^r=left(r^2times r-dfrac13(r)^3right)-left(r^2times0-dfrac13(0)^3right)$$ final answer is $dfrac43pi r^3$.
– Nosrati
Jul 20 at 21:26
Substitute $r$ instead of $x$.
– Nosrati
Jul 20 at 21:14
Substitute $r$ instead of $x$.
– Nosrati
Jul 20 at 21:14
At first upper bound $colorredminus$ the second lower bound!
– Nosrati
Jul 20 at 21:17
At first upper bound $colorredminus$ the second lower bound!
– Nosrati
Jul 20 at 21:17
$$left(r^2x-dfrac13x^3right)_0^r=left(r^2times r-dfrac13(r)^3right)-left(r^2times0-dfrac13(0)^3right)$$ final answer is $dfrac43pi r^3$.
– Nosrati
Jul 20 at 21:26
$$left(r^2x-dfrac13x^3right)_0^r=left(r^2times r-dfrac13(r)^3right)-left(r^2times0-dfrac13(0)^3right)$$ final answer is $dfrac43pi r^3$.
– Nosrati
Jul 20 at 21:26
add a comment |Â
3 Answers
3
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The volume is $$int_-r^r pi(r^2-x^2)dx=[pi(r^2x-tfrac13x^3)]_-r^r=tfrac4pi3r^3.$$Since the integrand is even, we could alternatively write $$int_-r^r pi(r^2-x^2)dx=int_0^r 2pi(r^2-x^2)dx=[2pi(r^2x-tfrac13x^3)]_0^r=tfrac4pi3r^3.$$
answered Jul 20 at 21:24
J.G.
13.2k11424
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0
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$$F(x)|_x=0^x=r=F(r)-F(0)$$
(by definition of the left term)
$$f(-x)=f(x) implies int_-r^rf(x) dx=2int_0^rf(x) dx$$
(consider the change of the lower bound too!)
answered Jul 20 at 21:25
Dr. Richard Klitzing
7586
add a comment |Â
up vote
0
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The volume of sphere is found by $$ V=2int _0^r pi (r^2-x^2) dx= 2pi int _0^r (r^2-x^2) dx$$
When you integrate you get $$r^2x -(1/3) x^3$$ and when you evaluate from $0$ to $r$ you get $$(2/3) r^3$$
Thus the total volume will be $$V= (4/3) pi r^3 $$
answered Jul 20 at 21:46
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The volume is $$int_-r^r pi(r^2-x^2)dx=[pi(r^2x-tfrac13x^3)]_-r^r=tfrac4pi3r^3.$$Since the integrand is even, we could alternatively write $$int_-r^r pi(r^2-x^2)dx=int_0^r 2pi(r^2-x^2)dx=[2pi(r^2x-tfrac13x^3)]_0^r=tfrac4pi3r^3.$$
answered Jul 20 at 21:24
J.G.
13.2k11424
add a comment |Â
up vote
1
down vote
The volume is $$int_-r^r pi(r^2-x^2)dx=[pi(r^2x-tfrac13x^3)]_-r^r=tfrac4pi3r^3.$$Since the integrand is even, we could alternatively write $$int_-r^r pi(r^2-x^2)dx=int_0^r 2pi(r^2-x^2)dx=[2pi(r^2x-tfrac13x^3)]_0^r=tfrac4pi3r^3.$$
answered Jul 20 at 21:24
J.G.
13.2k11424
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The volume is $$int_-r^r pi(r^2-x^2)dx=[pi(r^2x-tfrac13x^3)]_-r^r=tfrac4pi3r^3.$$Since the integrand is even, we could alternatively write $$int_-r^r pi(r^2-x^2)dx=int_0^r 2pi(r^2-x^2)dx=[2pi(r^2x-tfrac13x^3)]_0^r=tfrac4pi3r^3.$$
answered Jul 20 at 21:24
J.G.
13.2k11424
The volume is $$int_-r^r pi(r^2-x^2)dx=[pi(r^2x-tfrac13x^3)]_-r^r=tfrac4pi3r^3.$$Since the integrand is even, we could alternatively write $$int_-r^r pi(r^2-x^2)dx=int_0^r 2pi(r^2-x^2)dx=[2pi(r^2x-tfrac13x^3)]_0^r=tfrac4pi3r^3.$$
answered Jul 20 at 21:24
J.G.
13.2k11424
answered Jul 20 at 21:24
J.G.
13.2k11424
answered Jul 20 at 21:24
J.G.
13.2k11424
answered Jul 20 at 21:24
J.G.
13.2k11424
13.2k11424
add a comment |Â
add a comment |Â
up vote
0
down vote
$$F(x)|_x=0^x=r=F(r)-F(0)$$
(by definition of the left term)
$$f(-x)=f(x) implies int_-r^rf(x) dx=2int_0^rf(x) dx$$
(consider the change of the lower bound too!)
answered Jul 20 at 21:25
Dr. Richard Klitzing
7586
add a comment |Â
up vote
0
down vote
$$F(x)|_x=0^x=r=F(r)-F(0)$$
(by definition of the left term)
$$f(-x)=f(x) implies int_-r^rf(x) dx=2int_0^rf(x) dx$$
(consider the change of the lower bound too!)
answered Jul 20 at 21:25
Dr. Richard Klitzing
7586
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$F(x)|_x=0^x=r=F(r)-F(0)$$
(by definition of the left term)
$$f(-x)=f(x) implies int_-r^rf(x) dx=2int_0^rf(x) dx$$
(consider the change of the lower bound too!)
answered Jul 20 at 21:25
Dr. Richard Klitzing
7586
$$F(x)|_x=0^x=r=F(r)-F(0)$$
(by definition of the left term)
$$f(-x)=f(x) implies int_-r^rf(x) dx=2int_0^rf(x) dx$$
(consider the change of the lower bound too!)
answered Jul 20 at 21:25
Dr. Richard Klitzing
7586
answered Jul 20 at 21:25
Dr. Richard Klitzing
7586
answered Jul 20 at 21:25
Dr. Richard Klitzing
7586
answered Jul 20 at 21:25
Dr. Richard Klitzing
7586
7586
add a comment |Â
add a comment |Â
up vote
0
down vote
The volume of sphere is found by $$ V=2int _0^r pi (r^2-x^2) dx= 2pi int _0^r (r^2-x^2) dx$$
When you integrate you get $$r^2x -(1/3) x^3$$ and when you evaluate from $0$ to $r$ you get $$(2/3) r^3$$
Thus the total volume will be $$V= (4/3) pi r^3 $$
answered Jul 20 at 21:46
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
0
down vote
The volume of sphere is found by $$ V=2int _0^r pi (r^2-x^2) dx= 2pi int _0^r (r^2-x^2) dx$$
When you integrate you get $$r^2x -(1/3) x^3$$ and when you evaluate from $0$ to $r$ you get $$(2/3) r^3$$
Thus the total volume will be $$V= (4/3) pi r^3 $$
answered Jul 20 at 21:46
Mohammad Riazi-Kermani
27.5k41852
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The volume of sphere is found by $$ V=2int _0^r pi (r^2-x^2) dx= 2pi int _0^r (r^2-x^2) dx$$
When you integrate you get $$r^2x -(1/3) x^3$$ and when you evaluate from $0$ to $r$ you get $$(2/3) r^3$$
Thus the total volume will be $$V= (4/3) pi r^3 $$
answered Jul 20 at 21:46
Mohammad Riazi-Kermani
27.5k41852
The volume of sphere is found by $$ V=2int _0^r pi (r^2-x^2) dx= 2pi int _0^r (r^2-x^2) dx$$
When you integrate you get $$r^2x -(1/3) x^3$$ and when you evaluate from $0$ to $r$ you get $$(2/3) r^3$$
Thus the total volume will be $$V= (4/3) pi r^3 $$
answered Jul 20 at 21:46
Mohammad Riazi-Kermani
27.5k41852
answered Jul 20 at 21:46
Mohammad Riazi-Kermani
27.5k41852
answered Jul 20 at 21:46
Mohammad Riazi-Kermani
27.5k41852
answered Jul 20 at 21:46
Mohammad Riazi-Kermani
27.5k41852
27.5k41852
add a comment |Â
add a comment |Â
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Substitute $r$ instead of $x$.
– Nosrati
Jul 20 at 21:14
At first upper bound $colorredminus$ the second lower bound!
– Nosrati
Jul 20 at 21:17
$$left(r^2x-dfrac13x^3right)_0^r=left(r^2times r-dfrac13(r)^3right)-left(r^2times0-dfrac13(0)^3right)$$ final answer is $dfrac43pi r^3$.
– Nosrati
Jul 20 at 21:26