Recitation 11

Section 12.2 Problem 32: Find the volume of the solid by subtracting two volumes. The solid enclosed by the parabolic cylinder y = x^2 and the planes z = 3y, z = 2 + y.

Comment: The hardest part in this problem is to figure out how the solid looks like. Especially, which section of the solid we should choose as the base, that is the region of the double integration. It turns out that the section given by x=0 is the most convenient one.

Section 12.2 Problem 48: Evaluate the integral by reversing the order of integration. \int_0^8\int_{\sqrt[3]{y}}^2e^{x^4}dx dy

Comment: After reversing the order of integration, the integration becomes \int_0^2\int_0^{x^3}e^{x^4}dydx.

Section 12.2 Problem 57: Use geometry or symmetry, or both, to evaluate the double integral. \iint_D(2x+3y)dA, D is the rectangle 0\leq x\leq a, 0\leq y\leq b.

Solution: The integral is just the volume of the polyhedron enclosed by x=0, x=a, y=0, y=b, z=2x+3y. By symmetry, the volume is just half of the rectangular prism enclosed by x=0, x=a, y=0, y=b, z=0, z=2a+3b. Hence, the answer is \frac{1}{2}ab(2a+3b).

Section 12.3 Problem 9: Evaluate the given integral by changing to polar coordinates. \iint_R\sin(x^2+y^2)dA, where R is the region in the first quadrant between the circles with center the origin and radii 1 and 3.

Comment: Once we change the double integral to polar coordinates, we get \int_0^{\pi/2}\int_1^3\sin(r^2)rdrd\theta.

Section 12.3 Problem 19: Use polar coordinates to find the volume of the given solid. Inside both the cylinder x^2+y^2=4 and the ellipsoid 4x^2+4y^2+z^2=64.

Comment: Using the polar coordinates, x=r\cos\theta, y=r\sin\theta, the volume can be expressed as \int_0^{2\pi}\int_0^2 2\sqrt{64-4r^2}rdrd\theta.

Section 12.3 Problem 23: Evaluate the iterated integral by converting to polar coordinates. \int_{-3}^3\int_0^{\sqrt{9-x^2}}\sin(x^2+y^2)dydx

Comment: The integral becomes \int_0^\pi\int_0^3\sin(r^2)rdrd\theta under the polar coordinates.

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