Web$\begingroup$ alright I plugged in the right parametric values and my radical came out to be 1/4 and the whole thing came out to be 512/5 which is 102.4.. but the right answer is way bigger...??? $\endgroup$ WebJun 29, 2015 · $(x^2+y^2)dx−2xydy=0$ $\frac{dy}{dx}=\frac{x^2+y^2}{2xy} $..(i) This is a homogeneous differential equation because it has homogeneous functions of same degree 2. homogeneous functions are: $(x^2+y^2)$ and $2xy$, both functions have degree 2. Solution of differential equation: Equation (i) can be written as,
calculate the double integral (x^2+y^2)dxdy in the circle x^2 ... - Wyzant
WebUse Green’s Theorem to evaluate the line integral along the given positively oriented curve. integral C y^3dx-x^3dy, C is the circle x^2+y^2=4 Use Green’s Theorem to evaluate the line integral along the given positively oriented curve. ∫c cos y dx + x^2 sin y dy, C is the rectangle with vertices (0, 0), (5, 0), (5, 2), and (0, 2) WebPrecalculus. Find the Center and Radius x^2+y^2=4. x2 + y2 = 4 x 2 + y 2 = 4. This is the form of a circle. Use this form to determine the center and radius of the circle. (x−h)2 +(y−k)2 = r2 ( x - h) 2 + ( y - k) 2 = r 2. Match the values in this circle to those of the standard form. The variable r r represents the radius of the circle, h ... エジプシャンマウ オス 体重
Evaluate the line integral $\\int_C xy^4 ds $ of a half circle
WebJan 31, 2024 · C 5y3 dx ? 5x3 dy Use Green's Theorem to evaluate the line integral C is the circle x2 + y2 = 4 See answer Is the question mark supposed to be a plus or minus? Advertisement Advertisement LammettHash LammettHash ... cθ Select the correct answer below: −sinθ 1 sinθ −1 WebUse Green's Theorem to evaluate the line integral along the given positively oriented curve. 7y3 dx - 7x3 dy C is the circle x2 + y2 = 4 This problem has been solved! You'll get a … WebC −2y3 dx+2x3 dy where C is the circle of radius 3 centered at the origin. ANSWER: Using Green’s theorem we need to describe the interior of the region in order to set up the bounds for our double integral. This is best described with polar coordinates, 0 ≤ θ ≤ 2π and 0 ≤ r ≤ 3. And we get I C −2y3 dx+2x3 dy = ZZ D (6x2 +6y2)dA ... pancreatite incipiente