Chapter 13: Problem 6
In the integral for the moment \(M_{x z}\) with respect to the \(x z\) -plane of a solid, why does \(y\) appear in the integrand?
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Chapter 13: Problem 6
In the integral for the moment \(M_{x z}\) with respect to the \(x z\) -plane of a solid, why does \(y\) appear in the integrand?
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Use spherical coordinates to find the volume of the following solids. A ball of radius \(a>0\)
Linear transformations Consider the linear transformation \(T\) in \(\mathbb{R}^{2}\) given by \(x=a u+b v, y=c u+d v,\) where \(a, b, c,\) and \(d\) are real numbers, with \(a d \neq b c\) a. Find the Jacobian of \(T\) b. Let \(S\) be the square in the \(u v\) -plane with vertices (0,0) \((1,0),(0,1),\) and \((1,1),\) and let \(R=T(S) .\) Show that \(\operatorname{area}(R)=|J(u, v)|\) c. Let \(\ell\) be the line segment joining the points \(P\) and \(Q\) in the uv- plane. Show that \(T(\ell)\) (the image of \(\ell\) under \(T\) ) is the line segment joining \(T(P)\) and \(T(Q)\) in the \(x y\) -plane. (Hint: Use vectors.) d. Show that if \(S\) is a parallelogram in the \(u v\) -plane and \(R=T(S),\) then \(\operatorname{area}(R)=|J(u, v)| \operatorname{area}(S) .\) (Hint: Without loss of generality, assume the vertices of \(S\) are \((0,0),(A, 0)\) \((B, C),\) and \((A+B, C),\) where \(A, B,\) and \(C\) are positive, and use vectors.)
A cake is shaped like a hemisphere of radius 4 with its base on the \(x y\) -plane. A wedge of the cake is removed by making two slices from the center of the cake outward, perpendicular to the \(x y\) -plane and separated by an angle of \(\varphi\) a. Use a double integral to find the volume of the slice for \(\varphi=\pi / 4 .\) Use geometry to check your answer. b. Now suppose the cake is sliced by a plane perpendicular to the \(x y\) -plane at \(x=a > 0 .\) Let \(D\) be the smaller of the two pieces produced. For what value of \(a\) is the volume of \(D\) equal to the volume in part (a)?
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).
Use spherical coordinates to find the volume of the following solids. The solid bounded by the cylinders \(r=1\) and \(r=2,\) and the cones \(\varphi=\pi / 6\) and \(\varphi=\pi / 3\)
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