answer:To find the volume form, we first need to find the volume element dV. The volume element is given by the square root of the determinant of the metric tensor. The metric tensor g_{ij} is a 2times 2 symmetric matrix with components: g_{11} = r^2, g_{12} = g_{21} = 0, g_{22} = r^2sin^2(theta) The metric tensor's determinant is then: g = det(g_{ij}) = r^4sin^2(theta) Taking the square root, we get the volume element: dV = sqrt{g} dtheta dphi = r^2sin(theta) dtheta dphi Therefore, the volume form is: boxed{mathrm{Volume Form} = r^2sin(theta) dtheta dphi},To find the volume form of the given metric on the sphere, we need to compute the determinant of the metric tensor and then take the square root of the absolute value of the determinant. The metric tensor g_{ij} for the given metric is: g_{ij} = begin{pmatrix} r^2 & 0 0 & r^2sin^2(theta) end{pmatrix} Now, we compute the determinant of the metric tensor: det(g_{ij}) = detbegin{pmatrix} r^2 & 0 0 & r^2sin^2(theta) end{pmatrix} = r^4sin^2(theta) Next, we take the square root of the absolute value of the determinant: sqrt{|det(g_{ij})|} = sqrt{r^4sin^2(theta)} = r^2sin(theta) Finally, the volume form is given by the product of the square root of the determinant and the differential forms of the coordinates: dV = sqrt{|det(g_{ij})|} , dtheta wedge dphi = r^2sin(theta) , dtheta wedge dphi Since the radius of the sphere is given as r=3, we substitute this value into the volume form: dV = 3^2sin(theta) , dtheta wedge dphi = 9sin(theta) , dtheta wedge dphi So, the volume form of the given metric on the sphere is: dV = 9sin(theta) , dtheta wedge dphi,To find the volume form of the given metric on the sphere, we first need to find the determinant of the metric tensor. The metric tensor for the given metric is given by: g = begin{pmatrix} r^2 & 0 0 & r^2sin^2(theta) end{pmatrix} The determinant of the metric tensor is: det(g) = r^4 sin^2(theta) Now, we can compute the volume form, which is the square root of the determinant of the metric tensor times the differential volume element in the coordinate system: dV = sqrt{det(g)} , dtheta wedge dphi = r^2 sin(theta) , dtheta wedge dphi Since the radius of the sphere is given as r=3, the volume form on the sphere is: dV = 3^2 sin(theta) , dtheta wedge dphi = 9sin(theta) , dtheta wedge dphi

answer:To compute the Lie derivative of the vector field X with respect to the vector field Y, we use the Lie bracket X, Y. The Lie bracket is defined as X, Y = XY - YX, where XY and YX represent the action of the vector fields X and Y on functions. In this case, we have: X, Y = XY - YX First, let's compute XY: XY = (x ∂/∂x + y ∂/∂y)(2x ∂/∂x + y ∂/∂y) = x(-2 ∂/∂x + 0 ∂/∂y) + y(0 ∂/∂x + 1 ∂/∂y) = (-2x ∂/∂x + y ∂/∂y) Now, let's compute YX: YX = (2x ∂/∂x + y ∂/∂y)(x ∂/∂x + y ∂/∂y) = 2x(1 ∂/∂x + 0 ∂/∂y) + y(0 ∂/∂x + 1 ∂/∂y) = (2x ∂/∂x + y ∂/∂y) Now, we can compute the Lie bracket: X, Y = XY - YX = (-2x ∂/∂x + y ∂/∂y) - (2x ∂/∂x + y ∂/∂y) = -4x ∂/∂x + 2y ∂/∂y Finally, we can evaluate the Lie derivative at the point (1, 3): X, Y(1, 3) = -4(1) , To compute the Lie derivative of a vector field with respect to another vector field, we can use the following formula: L_Y(X) = Y,X = Y(X) - X(Y) where Y,X denotes the Lie bracket of the two vector fields. In this case, we have: X = x ∂/∂x + y ∂/∂y Y = 2x ∂/∂x + y ∂/∂y First, let's compute Y(X): Y(X) = (2x ∂/∂x + y ∂/∂y)(x ∂/∂x + y ∂/∂y) = 2x * (∂/∂x)(x) * ∂/∂x + 2x * (∂/∂x)(y) * ∂/∂y + y * (∂/∂y)(x) * ∂/∂x + y * (∂/∂y)(y) * ∂/∂y = 2x * 1 * ∂/∂x + 2x * 0 * ∂/∂y + y * 0 * ∂/∂x + y * 1 * ∂/∂y = 2x ∂/∂x + y ∂/∂y Next, let's compute X(Y): X(Y) = (x ∂/∂x + y ∂/∂y)(2x ∂/∂x + y ∂/∂y) = x * (∂/∂x)(2x) * ∂/∂x + x * (∂/∂x)(y) * ∂/∂y + y * (∂/∂y)(2x) * ∂/∂x + y * (∂/∂y)(y) * ∂/∂y = x * 2 * ∂/∂x + x * 0 * ∂/∂y + y * 0 * ∂/∂x + y * 1 * ∂/∂y = 2x ∂/∂x + y ∂/∂y Now, we can find the Lie derivative L_Y(X): L_Y(X) = Y(X) - X(Y) = (2x ∂/∂x + y ∂/∂y) - (2x ∂/∂x + y ∂/∂y) = 0 So, the Lie derivative of the vector field X with respect to the vector field Y at the point (1,3) on M is 0.,To compute the Lie derivative of the vector field X with respect to the vector field Y, we use the Lie bracket Y, X, which is defined as Y, X = YX - XY. First, let's compute YX: YX = (2x ∂/∂x + y ∂/∂y)(x ∂/∂x + y ∂/∂y) = 2x ∂/∂x (x ∂/∂x + y ∂/∂y) + y ∂/∂y (x ∂/∂x + y ∂/∂y) = 2x (x ∂²/∂x² + y ∂²/∂x∂y) + y (x ∂²/∂x∂y + y ∂²/∂y²) Now, let's compute XY: XY = (x ∂/∂x + y ∂/∂y)(2x ∂/∂x + y ∂/∂y) = x ∂/∂x (2x ∂/∂x + y ∂/∂y) + y ∂/∂y (2x ∂/∂x + y ∂/∂y) = x (2x ∂²/∂x² + y ∂²/∂x∂y) + y (2x ∂²/∂x∂y + y ∂²/∂y²) Now we compute the Lie bracket Y, X = YX - XY: Y, X = (2x (x ∂²/∂x² + y ∂²/∂x∂y) + y (x ∂²/∂x∂y + y ∂²/∂y²)) - (x (2x ∂²/∂x² + y ∂²/∂x∂y) + y (2x ∂²/∂x∂y + y ∂²/∂y²)) = x ∂²/∂x² + y ∂²/∂x∂y Now, we evaluate the Lie derivative at the point (1, 3): L_YX(1, 3) = 1 ∂²/∂x² + 3 ∂²/∂x∂y Since the Lie derivative is a vector field, the result is still a vector field. In this case, the Lie derivative of X with respect to Y at the point (1, 3) is given by the vector field L_YX = ∂/∂x + 3 ∂/∂y.

answer:To compute the Lie derivative of vector field V along vector field Y, we use the Lie bracket Y, V. The Lie bracket is defined as Y, V = Y(V) - V(Y), where Y(V) and V(Y) are the action of Y on V and the action of V on Y, respectively. First, let's compute Y(V): Y(V) = (2x∂/∂x + ∂/∂y)(x^2 ∂/∂x + x∂/∂y) = 2x∂/∂x(x^2 ∂/∂x + x∂/∂y) + ∂/∂y(x^2 ∂/∂x + x∂/∂y) Now, apply the product rule: Y(V) = 2x(2x ∂/∂x + x ∂/∂y) + (2x^2 ∂/∂y + x ∂/∂x) Next, compute V(Y): V(Y) = (x^2 ∂/∂x + x∂/∂y)(2x∂/∂x + ∂/∂y) = x^2 ∂/∂x(2x ∂/∂x + ∂/∂y) + x∂/∂y(2x ∂/∂x + ∂/∂y) Apply the product rule: V(Y) = x^2(2 ∂/∂x + x ∂/∂y) + x(2x ∂/∂y + ∂/∂x) Now, compute the Lie bracket Y, V = Y(V) - V(Y): Y, V = (2x(2x ∂/∂x + x ∂/∂y) + (2x^2 ∂/∂y + x ∂/∂x)) - (x^2(2 ∂/∂x + x ∂/∂y) + x(2x ∂/∂y + ∂/∂x)) Simplify the expression: Y, V = 2x^2 ∂/∂x + x^2 ∂/∂y Finally, evaluate the Lie derivative at the point (2, 3): L_Y(V)(2, 3) = (2(2)^2 ∂/∂x + (2)^2 ∂/∂y) = 8 ∂/∂x + 4 ∂/∂y Thus, the Lie derivative of V along Y at the point (2, 3) is 8 ∂/∂x + 4 ∂/∂y.,To compute the Lie derivative of a vector field V along another vector field Y, we use the Lie bracket Y, V. The Lie bracket is defined as: Y, V = YV - VY In our case, Y = 2x∂/∂x + ∂/∂y and V = x^2 ∂/∂x + x∂/∂y. Now, let's compute the Lie bracket: Y, V = YV - VY = (2x∂/∂x + ∂/∂y)(x^2 ∂/∂x + x∂/∂y) - (x^2 ∂/∂x + x∂/∂y)(2x∂/∂x + ∂/∂y) First, compute YV: (2x∂/∂x + ∂/∂y)(x^2 ∂/∂x + x∂/∂y) = 2x(x^2 ∂/∂x + x∂/∂y)∂/∂x + (x^2 ∂/∂x + x∂/∂y)∂/∂y = 2x(x^3 ∂/∂x^2 + 2x^2 ∂/∂x) + (x^3 ∂/∂x + x^2 ∂/∂y) Next, compute VY: (x^2 ∂/∂x + x∂/∂y)(2x∂/∂x + ∂/∂y) = x^2(2x∂/∂x + ∂/∂y)∂/∂x + x(2x∂/∂x + , To compute the Lie derivative of a vector field V with respect to another vector field Y, we use the following formula: L_Y V = Y, V where Y, V is the Lie bracket of Y and V, defined as: Y, V = Y(V^i) ∂/∂x^i - V(Y^i) ∂/∂x^i Here, V^i and Y^i are the components of the vector fields V and Y, respectively. In our case, we have: V = x^2 ∂/∂x + x∂/∂y => V^1 = x^2, V^2 = x Y = 2x∂/∂x + ∂/∂y => Y^1 = 2x, Y^2 = 1 Now let's compute the derivatives in the Lie bracket: Y(V^1) = Y(x^2) = 2x * d(x^2)/dx = 4x^2 Y(V^2) = Y(x) = d(x)/dx = 1 V(Y^1) = V(2x) = 2 * d(2x)/dx = 2 V(Y^2) = V(1) = d(1)/dx = 0 Now we can assemble the Lie bracket: Y, V = (4x^2 - 2) ∂/∂x + (1 - 0) ∂/∂y To find the Lie derivative of V along Y at the point (2, 3), we need to evaluate this Lie bracket at that point: Y, V|_{(2, 3)} = (4(2)^2 - 2) ∂/∂x + (1 - 0) ∂/∂y = 14 ∂/∂x + 1 ∂/∂y So the Lie derivative of V along Y at the point (2, 3) is given by the vector (14, 1).

answer:To compute the Lie derivative of X with respect to Y, we need to calculate the Lie bracket Y, X. Recall that the Lie bracket of two vector fields is defined as the commutator of their actions on smooth functions, i.e., Y, X(f) = Y(X(f)) - X(Y(f)) for any smooth function f on M. In local coordinates, we can express the vector fields X and Y as: X = x^i frac{partial}{partial x^i}, quad Y = y^i frac{partial}{partial x^i}, where the Einstein summation convention is used, and x^i and y^i are the components of X and Y in the local coordinates, respectively. Now, let's compute the Lie bracket Y, X: Y, X = Y(X^i frac{partial}{partial x^i}) - X(Y^i frac{partial}{partial x^i}). Applying the product rule for differentiation, we get: Y, X = (Y^j frac{partial x^i}{partial x^j}) frac{partial}{partial x^i} + x^i (Y^j frac{partial}{partial x^j} frac{partial}{partial x^i}) - (X^j frac{partial y^i}{partial x^j}) frac{partial}{partial x^i} - y^i (X^j frac{partial}{partial x^j} frac{partial}{partial x^i}). Now, we can simplify the expression: Y, X = (Y^j frac{partial x^i}{partial x^j} - X^j frac{partial y^i}{partial x^j}) frac{partial}{partial x^i}. Thus, the Lie derivative of X with respect to Y in local coordinates is given by: Y, X = (Y^j frac{partial x^i}{partial x^j} - X^j frac{partial y^i}{partial x^j}) frac{partial}{partial x^i}. ,To compute the Lie derivative of X with respect to Y, we need to calculate the Lie bracket Y, X. In local coordinates, the components of the Lie bracket Y, X are given by the partial derivatives of the components of the vector fields X with respect to the components of Y: Y, X^i = frac{partial x^i}{partial y^j} cdot Y^j - frac{partial x^i}{partial y^j} cdot X^j, where i = 1, 2, ..., n, and j = 1, 2, ..., n. So, to determine the Lie derivative of X with respect to Y, we need to compute the partial derivatives frac{partial x^i}{partial y^j}, and then apply the formula above., To calculate the Lie derivative of X with respect to Y, denoted by Y,X, we can use the following formula in local coordinates: Y,X = left( y^j frac{partial x^i}{partial x^j} - x^j frac{partial y^i}{partial x^j} right) frac{partial}{partial x^i} Here, x^i and y^i are the components of the vector fields X and Y respectively in the local coordinates x^1, x^2, ..., x^n on M. Now let's compute each term in the above formula. First, note that y^j frac{partial x^i}{partial x^j} = y^j delta_j^i = y^i where delta_j^i is the Kronecker delta. Next, we have begin{align*} x^j frac{partial y^i}{partial x^j} &= x^j frac{partial}{partial x^j} left( y^i(x^1, ..., x^n) right) &= x^j sum_{k=1}^n frac{partial y^i}{partial x^k} frac{partial x^k}{partial x^j} &= sum_{j=1}^n x^j sum_{k=1}^n frac{partial y^i}{partial x^k} delta_j^k &= sum_{j=1}^n x^j frac{partial y^i}{partial x^j} end{align*} Now we can calculate the Lie derivative: begin{align*} Y,X &= left( y^j frac{partial x^i}{partial x^j} - x^j frac{partial y^i}{partial x^j} right) frac{partial}{partial x^i} &= left( y^i - sum_{j=1}^n x^j frac{partial y^i}{partial x^j} right) frac{partial}{partial x^i} end{align*} So the components of the Lie derivative Y,X in these local coordinates are given by Y,X^i = y^i - sum_{j=1}^n x^j frac{partial y^i}{partial x^j}.