Chapter 13: Problem 62
Reverse the order of integration in the following integrals. $$\int_{1}^{e} \int_{0}^{\ln x} f(x, y) d y d x$$
Chapter 13: Problem 62
Reverse the order of integration in the following integrals. $$\int_{1}^{e} \int_{0}^{\ln x} f(x, y) d y d x$$
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Get started for freeEvaluate the following integrals using the method of your choice. A sketch is helpful. $$\iint_{R} \sqrt{x^{2}+y^{2}} d A ; R=\left\\{(x, y): 1 \leq x^{2}+y^{2} \leq 4\right\\}$$
An important integral in statistics associated with the normal distribution is \(I=\int_{-\infty}^{\infty} e^{-x^{2}} d x .\) It is evaluated in the following steps. a. Assume that $$\begin{aligned} I^{2} &=\left(\int_{-\infty}^{\infty} e^{-x^{2}} d x\right)\left(\int_{-\infty}^{\infty} e^{-y^{2}} d y\right) \\ &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}-y^{2}} d x d y \end{aligned}$$ where we have chosen the variables of integration to be \(x\) and \(y\) and then written the product as an iterated integral. Evaluate this integral in polar coordinates and show that \(I=\sqrt{\pi} .\) Why is the solution \(I=-\sqrt{\pi}\) rejected? b. Evaluate \(\int_{0}^{\infty} e^{-x^{2}} d x, \int_{0}^{\infty} x e^{-x^{2}} d x,\) and \(\int_{0}^{\infty} x^{2} e^{-x^{2}} d x\) (using part (a) if needed).
Find equations for the bounding surfaces, set up a volume integral, and evaluate the integral to obtain a volume formula for each region. Assume that \(a, b, c, r, R,\) and h are positive constants. Find the volume of the cap of a sphere of radius \(R\) with height \(h\)
In Section 10.3 it was shown that the area of a region enclosed by the polar curve \(r=g(\theta)\) and the rays \(\theta=\alpha\) and \(\theta=\beta,\) where \(\beta-\alpha \leq 2 \pi,\) is \(A=\frac{1}{2} \int_{\alpha}^{\beta} r^{2} d \theta .\) Prove this result using the area formula with double integrals.
Find the coordinates of the center of mass of the following solids with variable density. The interior of the prism formed by \(z=x, x=1, y=4,\) and the coordinate planes with \(\rho(x, y, z)=2+y\)
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