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Let’s check out your problem:
Use Green's theorem to evaluate the line integral along the given positively oriented curve,
∫
C
7
y
3
d
x
−
7
x
3
d
y
,
is the circle
x
2
+
y
2
=
4
\int_{C}7y^{3}\,dx-7x^{3}\,dy,\quad \text{is the circle } x^{2}+y^{2}=4
∫
C
7
y
3
d
x
−
7
x
3
d
y
,
is the circle
x
2
+
y
2
=
4
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Math Problems
Calculus
Find indefinite integrals using the substitution
Full solution
Q.
Use Green's theorem to evaluate the line integral along the given positively oriented curve,
∫
C
7
y
3
d
x
−
7
x
3
d
y
,
is the circle
x
2
+
y
2
=
4
\int_{C}7y^{3}\,dx-7x^{3}\,dy,\quad \text{is the circle } x^{2}+y^{2}=4
∫
C
7
y
3
d
x
−
7
x
3
d
y
,
is the circle
x
2
+
y
2
=
4
Apply Green's Theorem:
Use Green's theorem to convert the line integral into a double integral over the region enclosed by the curve.
Calculate Partial Derivatives:
Calculate the partial derivatives
∂
Q
∂
x
\frac{\partial Q}{\partial x}
∂
x
∂
Q
and
∂
P
∂
y
\frac{\partial P}{\partial y}
∂
y
∂
P
.
Substitute into Double Integral:
Substitute the partial derivatives into the double integral.
Convert to Polar Coordinates:
Recognize the region
R
R
R
as the disk
x
2
+
y
2
≤
4
x^2 + y^2 \leq 4
x
2
+
y
2
≤
4
. Convert to polar coordinates where
x
=
r
cos
(
θ
)
x = r \cos(\theta)
x
=
r
cos
(
θ
)
and
y
=
r
sin
(
θ
)
y = r \sin(\theta)
y
=
r
sin
(
θ
)
.
Set up Limits for Integration:
Substitute polar coordinates into the integral and set up the limits for
r
r
r
and
θ
\theta
θ
.
Simplify Using Trigonometric Identity:
Simplify the integrand using the identity
cos
2
(
θ
)
+
sin
2
(
θ
)
=
1
\cos^2(\theta) + \sin^2(\theta) = 1
cos
2
(
θ
)
+
sin
2
(
θ
)
=
1
.
Integrate with Respect to
r
r
r
:
Integrate with respect to
r
r
r
first.
Integrate with Respect to
θ
\theta
θ
:
Integrate the result with respect to
θ
\theta
θ
.
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[
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\frac{d^{2}}{d x^{2}}\left[5 \ln \left(x^{5}\right)\right]
d
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[
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Consider the curve given by the equation
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d
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Consider the curve given by the equation
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2
x
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d
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(
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(\square
(
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Can this differential equation be solved using separation of variables?
\newline
d
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y
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y
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d
x
d
y
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2
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y
3
\newline
Choose
1
1
1
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\newline
(A) Yes
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Question
y
=
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4
d
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y
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4
d
x
d
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Solve the equation. Check your solution
\newline
19
=
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−
(
z
+
5
)
19=2-(z+5)
19
=
2
−
(
z
+
5
)
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The solution set is \(\{\longrightarrow\}. (Simplify your answer
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tan
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1
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d
t
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∫
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i
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t
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Write the answer in proper scientific notation:
\newline
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×
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0
5
×
1
0
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)
4
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(
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0
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Question
Solve for
x
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\newline
x
−
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<
0
\frac{x}{-2} < 0
−
2
x
<
0
\newline
Answer Attempt
1
1
1
out of
2
2
2
\newline
<
<
<
\newline
≥
\geq
≥
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≥
\geq
≥
or
\newline
Inequality Notation:
\newline
Number Line:
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−
∫
t
4
(
1
−
t
2
)
2
d
t
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−
∫
(
1
−
t
2
)
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t
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d
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