Cartesian to polar coordinates Sketch the given region of
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- No integrals Let F = ⟨2z, z, 2y + x⟩, and let S be the hemisphereof radius a with its base in the xy-plane and center at the origin.a. Evaluate ∫∫S (∇ x F) ⋅ n dS by computing ∇ x F and appealing to symmetry.b. Evaluate the line integral using Stokes’ Theorem to check part (a).arrow_forwardA solid S is generated by revolving the region between the x-axis and the curve y =√ sinx (0 ≤ x ≤ π) about the x-axis.(a) For x between 0 and π, the crosssectional area of S perpendicular to the xaxis at x is A(x) = _____.(b) An integral expression for the volume of S is _____ .(c) The value of the integral in part (b) is_____ .arrow_forwardThe region of integration R is given by the following inequalities.R : x^2 + y^2 ≤ 1, x + y ≥ 1Evaluate the integral below by switching to polar coordinates. ∫ ∫(x − y)dAarrow_forward
- Work through all integrals. Determine the volumes of the solids of revolution generated by revolving the given region about the given line. Do by the method indicated. - The region bounded by y = sin(x) , y = 0, on [0,pi], is revolved about the y = 1. Do by washers.arrow_forward4.Find first derivates y=(lnx)^lnx 5.Find the area of the region bounded by the graphs of x=3−y² and x=1+y. 6.Finding the Area of a Surface of Revolution. Set up and evaluate the definite integral forthe area of the surface generated by revolving the curve aboutthe y-axis. y=1−x²/4, 0≤x≤2arrow_forwardA thick spherical shell occupies the region between two spheres of radii a and 2a, both centred on the origin. The shell is made of a material with density p = A(x2 + y2) z2, where A is a constant. Hence, or otherwise, find the mass of the shell by evaluating a suitable volume integral.You may find the substitution u = cosθ useful.arrow_forward
- P10) The region R in the xy-plane is bounded by the circles x2 + y2 = 4 and x2 + (y − 2)2 = 4 I. Set up, but do not evaluate, an integral or sum of integrals in Cartesian coordinates that represents the area of R. You may choose the variable of integration freely. II. Set up, but do not evaluate, an integral or sum of integrals using polar coordinates that represents the area of R. See details in image uploaded please.arrow_forwardConverting to a polar integral Integrate ƒ(x, y) = [ln (x2 + y2 ) ]/(x2 + y2) over the region 1<= x2 + y2<= e^2.arrow_forward34) The figure shows the region of integration for the integral. Rewrite this itegral as an equivalent iterated integral in the five other orders. ∫0 to 1 ∫0 to (1-x^2) ∫0 to (1-x) f(x, y, z) dydzdxarrow_forward
- a) The figure shows the region of integration for the integral. 6 0 36−x2 0 6−x f(x, y, z) dy dz dx 0 Rewrite this integral as an equivalent iterated integral in the five other orders. Assume y(x) = 6 − x and z(x) = 36 − x2 b) Sketch the solid whose volume is given by the iterated integral. 1 0 1− x 0 9 − 9z dy dz dx 0arrow_forwardConsider the region bounded by the graphs of y = ln x, y = ex, x = 1, and x = e. Set up the definite integral that solves for the volume of the solid when the region is revolved about:a. the x-axis b. the y-axis c. the line x = -1 d. the line y = -1arrow_forwardSet up integrals for both orders of integration. Use the more convenient order to evaluate the integral over the plane region R. R sin x sin y dA R: rectangle with vertices (−?, 0), (?, 0), (?, ?/2), (−?, ?/2)arrow_forward
- Calculus: Early TranscendentalsCalculusISBN:9781285741550Author:James StewartPublisher:Cengage LearningThomas' Calculus (14th Edition)CalculusISBN:9780134438986Author:Joel R. Hass, Christopher E. Heil, Maurice D. WeirPublisher:PEARSONCalculus: Early Transcendentals (3rd Edition)CalculusISBN:9780134763644Author:William L. Briggs, Lyle Cochran, Bernard Gillett, Eric SchulzPublisher:PEARSON
- Calculus: Early TranscendentalsCalculusISBN:9781319050740Author:Jon Rogawski, Colin Adams, Robert FranzosaPublisher:W. H. FreemanCalculus: Early Transcendental FunctionsCalculusISBN:9781337552516Author:Ron Larson, Bruce H. EdwardsPublisher:Cengage Learning