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Math  /  Calculus

QuestionEvaluate: ∮(7y−esin⁡(x))dx+[15x−sin⁡(y3+8y)]dy\oint\left(7 y-e^{\sin (x)}\right) d x+\left[15 x-\sin \left(y^{3}+8 y\right)\right] d y using one method. CC is the boundary of the graph of a circle of radius 3 .

Studdy Solution

STEP 1

1. The problem involves evaluating a line integral over a closed curve C C .
2. The curve C C is the boundary of a circle of radius 3, centered at the origin.
3. We will use Green's Theorem to evaluate the line integral.

STEP 2

1. Verify the conditions for Green's Theorem.
2. Apply Green's Theorem to convert the line integral into a double integral.
3. Set up and evaluate the double integral over the region enclosed by C C .

STEP 3

Verify the conditions for Green's Theorem. Green's Theorem states that for a positively oriented, simple closed curve C C and a region R R it encloses, if F=M dx+N dy \mathbf{F} = M \, dx + N \, dy is continuously differentiable on an open region containing R R , then:
∮CM dx+N dy=∬R(∂N∂x−∂M∂y) dA\oint_C M \, dx + N \, dy = \iint_R \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \, dA
For our problem, M(x,y)=7y−esin⁡(x) M(x, y) = 7y - e^{\sin(x)} and N(x,y)=15x−sin⁡(y3+8y) N(x, y) = 15x - \sin(y^3 + 8y) .

STEP 4

Compute the partial derivatives needed for Green's Theorem:
∂N∂x=∂∂x(15x−sin⁡(y3+8y))=15\frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(15x - \sin(y^3 + 8y)) = 15
∂M∂y=∂∂y(7y−esin⁡(x))=7\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(7y - e^{\sin(x)}) = 7
Apply Green's Theorem:
∮C(7y−esin⁡(x)) dx+(15x−sin⁡(y3+8y)) dy=∬R(15−7) dA=∬R8 dA\oint_C (7y - e^{\sin(x)}) \, dx + (15x - \sin(y^3 + 8y)) \, dy = \iint_R (15 - 7) \, dA = \iint_R 8 \, dA

STEP 5

Set up and evaluate the double integral over the region R R , which is the interior of the circle of radius 3 centered at the origin:
The area dA dA in polar coordinates is r dr dθ r \, dr \, d\theta . The limits for r r are from 0 to 3, and for θ \theta are from 0 to 2π 2\pi .
∬R8 dA=∫02π∫038⋅r dr dθ\iint_R 8 \, dA = \int_0^{2\pi} \int_0^3 8 \cdot r \, dr \, d\theta
Evaluate the inner integral:
∫038r dr=8[r22]03=8⋅92=36\int_0^3 8r \, dr = 8 \left[ \frac{r^2}{2} \right]_0^3 = 8 \cdot \frac{9}{2} = 36
Now, evaluate the outer integral:
∫02π36 dθ=36⋅[θ]02π=36⋅2π=72π\int_0^{2\pi} 36 \, d\theta = 36 \cdot \left[ \theta \right]_0^{2\pi} = 36 \cdot 2\pi = 72\pi
The value of the line integral is:
72π \boxed{72\pi}

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