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Chapter 16: Problem 8
Evaluate the following iterated integrals. $$\int_{1}^{2} \int_{0}^{1}\left(3 x^{2}+4 y^{3}\right) d y d x$$
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Consider the surface \(z=x^{2}-y^{2}\) a. Find the region in the \(x y\) -plane in polar coordinates for which \(z \geq 0\) b. Let \(R=\\{(r, \theta): 0 \leq r \leq a,-\pi / 4 \leq \theta \leq \pi / 4\\},\) which is a sector of a circle of radius \(a\). Find the volume of the region below the hyperbolic paraboloid and above the region \(R\)
A triangular region has a base that connects the vertices (0,0) and \((b, 0),\) and a third vertex at \((a, h),\) where \(a>0, b>0,\) and \(h>0\) a. Show that the centroid of the triangle is \(\left(\frac{a+b}{3}, \frac{h}{3}\right)\) b. Recall that the three medians of a triangle extend from each vertex to the midpoint of the opposite side. Knowing that the medians of a triangle intersect in a point \(M\) and that each median bisects the triangle, conclude that the centroid of the triangle is \(M.\)
A cylindrical soda can has a radius of \(4 \mathrm{cm}\) and a height of \(12 \mathrm{cm} .\) When the can is full of soda, the center of mass of the contents of the can is \(6 \mathrm{cm}\) above the base on the axis of the can (halfway along the axis of the can). As the can is drained, the center of mass descends for a while. However, when the can is empty (filled only with air), the center of mass is once again \(6 \mathrm{cm}\) above the base on the axis of the can. Find the depth of soda in the can for which the center of mass is at its lowest point. Neglect the mass of the can, and assume the density of the soda is \(1 \mathrm{g} / \mathrm{cm}^{3}\) and the density of air is \(0.001 \mathrm{g} / \mathrm{cm}^{3} .\)
Find the coordinates of the center of mass of the following solids with variable density. The solid bounded by the upper half of the sphere \(\rho=6\) and \(z=0\) with density \(f(\rho, \varphi, \theta)=1+\rho / 4\)
The solid bounded by the paraboloid \(z=2-x^{2}-y^{2}\) and the plane \(z=1\).
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