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zvonat [6]
3 years ago
10

Evaluate.

Mathematics
1 answer:
madreJ [45]3 years ago
8 0

Answer:

<em>Answer</em><em> </em><em>is</em><em> </em><em>none</em><em> </em><em>of</em><em> </em><em>these</em><em>.</em>

<em>Answer</em><em> </em><em>is given below with explanations</em><em>. </em>

Step-by-step explanation:

<em>Given</em><em> </em><em>that</em><em> </em><em>x</em><em> </em><em>=</em><em> </em><em>-2</em><em>.</em><em>5</em>

<em>To</em><em> </em><em>find</em><em> </em><em>the</em><em> </em><em>value</em><em> </em><em>of</em><em> </em><em>x</em><em>+</em><em>13</em>

<em>Ob</em><em> </em><em>substituting</em><em> </em><em>the</em><em> </em><em>value</em><em> </em><em>in</em><em> </em><em>the</em><em> </em><em>given equation</em><em> </em>

<em>we</em><em> </em><em>get</em><em> </em>

<em>=</em><em> </em><em>-2</em><em>.</em><em>5</em><em>+</em><em>13</em>

<em>=</em><em> </em><em>13</em><em> </em><em>-</em><em> </em><em>2</em><em>.</em><em>5</em><em> </em><em> </em><em> </em><em> </em><em>(</em><em>we</em><em> </em><em>can</em><em> </em><em>also</em><em> </em><em>write</em><em> </em><em>like</em><em> </em><em>this</em><em>)</em>

<em>=</em><em> </em><em>10</em><em>.</em><em>5</em>

<em>Therefore</em><em> </em><em>value</em><em> </em><em>of</em><em> </em><em>x</em><em> </em><em>+</em><em>13</em><em> </em><em>is</em><em> </em><em>10</em><em>.</em><em>5</em>

<em>HAVE</em><em> </em><em>A NICE DAY</em><em>!</em>

<em>THANKS FOR GIVING ME THE OPPORTUNITY</em><em> </em><em>TO ANSWER YOUR QUESTION</em><em>. </em>

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Rina8888 [55]

The line integral along the given positively oriented curve is -216π. Using green's theorem, the required value is calculated.

<h3>What is green's theorem?</h3>

The theorem states that,

\int_CPdx+Qdy = \int\int_D(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dx dy

Where C is the curve.

<h3>Calculation:</h3>

The given line integral is

\int_C9y^3dx-9x^3dy

Where curve C is a circle x² + y² = 4;

Applying green's theorem,

P = 9y³; Q = -9x³

Then,

\frac{\partial P}{\partial y} = \frac{\partial 9y^3}{\partial y} = 27y^2

\frac{\partial Q}{\partial x} = \frac{\partial -9x^3}{\partial x} = 27x^2

\int_C9y^3dx-9x^3dy = \int\int_D(-27x^2 - 27y^2)dx dy

⇒ -27\int\int_D(x^2 + y^2)dx dy

Since it is given that the curve is a circle i.e., x² + y² = 2², then changing the limits as

0 ≤ r ≤ 2; and 0 ≤ θ ≤ 2π

Then the integral becomes

-27\int\limits^{2\pi}_0\int\limits^2_0r^2. r dr d\theta

⇒ -27\int\limits^{2\pi}_0\int\limits^2_0 r^3dr d\theta

⇒ -27\int\limits^{2\pi}_0 (r^4/4)|_0^2 d\theta

⇒ -27\int\limits^{2\pi}_0 (16/4) d\theta

⇒ -108\int\limits^{2\pi}_0 d\theta

⇒ -108[2\pi - 0]

⇒ -216π

Therefore, the required value is -216π.

Learn more about green's theorem here:

brainly.com/question/23265902

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