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Problem 12

Evaluate the double integral \(\int_{R} \int f(r, \theta) d A\), and sketch the region \(R\). $$\int_{0}^{\pi / 2} \int_{0}^{3} r e^{-r^{2}} d r d \theta$$

Problem 12

Use the indicated change of variables to evaluate the double integral. $$ \begin{aligned} &\int_{R} \int 60 x y d A \\ &x=\frac{1}{2}(u+v) \\ &y=-\frac{1}{2}(u-v) \end{aligned} $$

Problem 12

Find the mass and center of mass of the lamina bounded by the graphs of the equations for the given density or densities. (Hint: Some of the integrals are simpler in polar coordinates.) \(x^{2}+y^{2}=a^{2}, 0 \leq x, 0 \leq y\) (a) \(\rho=k\) (b) \(\rho=k\left(x^{2}+y^{2}\right)\)

Problem 13

Convert the integral from rectangular coordinates to both cylindrical and spherical coordinates, and evaluate the simplest iterated integral. $$ \int_{-2}^{2} \int_{-\sqrt{4-x^{2}}}^{\sqrt{4-x^{2}}} \int_{x^{2}+y^{2}}^{4} x d z d y d x $$

Problem 13

Set up a triple integral for the volume of the solid. The solid in the first octant bounded by the coordinate planes and the plane \(z=4-x-y\)

Problem 13

Evaluate the double integral \(\int_{R} \int f(r, \theta) d A\), and sketch the region \(R\). \(\int_{0}^{\pi / 2} \int_{0}^{1+\sin \theta} \theta r d r d \theta\)

Problem 13

Find the area of the surface given by \(z=f(x, y)\) over the region \(R .\) (Hint: Some of the integrals are simpler in polar coordinates.) $$ \begin{aligned} &f(x, y)=\sqrt{a^{2}-x^{2}-y^{2}} \\ &R=\left\\{(x, y): x^{2}+y^{2} \leq b^{2}, 0

Problem 13

Use the indicated change of variables to evaluate the double integral. $$ \begin{aligned} &\int_{R} \int y(x-y) d A \\ &x=u+v \\ &y=u \end{aligned} $$

Problem 13

Set up an integral for both orders of integration, and use the more convenient order to evaluate the integral over the region \(R\). $$\int_{R} \int x y d A$$ \(R:\) rectangle with vertices \((0,0),(0,5),(3,5),(3,0)\)

Problem 13

Find the mass and center of mass of the lamina bounded by the graphs of the equations for the given density or densities. (Hint: Some of the integrals are simpler in polar coordinates.) \(y=\sqrt{x}, y=0, x=4, \rho=k x y\)

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