Change the order of integration, but do not evaluate, the following integrals:
Change the order of integration and evaluate: \[ \int_{0}^{1} \int_{y}^{1} \sin (x^2) \, {\it dx}\, {\it dy}. \]
In the following integrals, change the order of integration, sketch the corresponding regions, and evaluate the integral both ways.
Find
Change the order of integration and evaluate: \[ \int_{0}^{1} \int_{\sqrt{y}}^{1} e^{x^{3}} \, {\it dx}\, {\it dy}. \]
Consider the intuitive fact that if a region \(D\) in \(\mathbb R^{2}\) can be split into a disjoint union of subsets \(D = D_1 \cup D_2\), then a double integral over \(D\) may also be divided into a sum of two integrals: \[ \intop\!\!\!\intop\nolimits_{D} f(x,y) \, {\it dA} = \intop\!\!\!\intop\nolimits_{D_{1}} f(x,y) \, {\it dA} + \intop\!\!\!\intop\nolimits_{D_{2}} f(x,y) \, {\it dA}. \] (See Section 15.2 for the analogous statement over a rectangular box.) Are the following attempts to change the order of integration true or false?
If \(f(x,y)=e^{\sin(x+y)}\) and \(D=[-\pi,\pi]\times [-\pi,\pi]\), show that \[ \frac{1}{e}\leq \frac{1}{4\pi^2}\intop\!\!\!\intop\nolimits_{D}\, f(x,y)\ {\it dA}\leq e. \]
Show that \[ \frac{1}{2}(1-\cos 1)\leq \intop\!\!\!\intop\nolimits_{[0,1]\times [0,1]} \frac{\sin x}{1+(xy)^4} {\it dx} \ {\it dy} \leq 1. \]
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If \(D=[-1,1]\times [-1,2]\), show that \[ 1\leq \intop\!\!\!\intop\nolimits_{D} \frac{{\it dx}\,{\it dy} }{x^2 +y^2+1}\leq 6. \]
Using the mean-value inequality, show that \[ \frac{1}{6}\leq \intop\!\!\!\intop\nolimits_{D}\, \frac{{\it dA}}{y-x+3}\leq \frac{1}{4}, \] where \(D\) is the triangle with vertices (0, 0), (1, 1), and (1, 0).
Compute the volume of an ellipsoid with semiaxes \(a, b\), and \(c\). (HINT: Use symmetry and first find the volume of one half of the ellipsoid.)
Compute \(\displaystyle\intop\!\!\!\intop\nolimits_Df(x,y)\, {\it dA}\), where \(f(x,y)=y^2\sqrt{x}\) and \(D\) is the set of \((x, y)\), where \(x>0\), \(y>x^2\), and \(y<10-x^2\).
Find the volume of the region determined by \(x^2+y^2+z^2\leq 10, z\geq 2\). Use the disk method from one-variable calculus and state how the method is related to Cavalieri’s principle.
Evaluate \(\displaystyle\intop\!\!\!\intop\nolimits_D e^{x-y} {\it dx}\, {\it dy}\), where \(D\) is the interior of the triangle with vertices (0, 0), (1, 3), and (2, 2).
Evaluate \(\displaystyle\intop\!\!\!\intop\nolimits_D y^3(x^2+y^2)^{-3/2} {\it dx}\, {\it dy}\), where \(D\) is the region determined by the conditions \(\frac{1}{2}\leq y\leq 1\) and \(x^2+y^2\leq 1\).
Given that the double integral \(\displaystyle\intop\!\!\!\intop\nolimits_D f(x,y)\, {\it dx}\, {\it dy}\) of a positive continuous function \(f\) equals the iterated integral \(\displaystyle\int_0^1\left[\int_{x^2}^xf(x,y)\, {\it dy} \right] {\it dx} \), sketch the region \(D\) and interchange the order of integration.
Given that the double integral \(\displaystyle\intop\!\!\!\intop\nolimits_Df(x,y)\, {\it dx}\, {\it dy}\) of a positive continuous function \(f\) equals the iterated integral \(\displaystyle\int_0^1\left[\int_y^{\sqrt{2-y^2}}f(x,y)\, {\it dx}\right] {\it dy} \), sketch the region \(D\) and interchange the order of integration.
Prove that \(\displaystyle 2\int_a^b\int_x^b f(x)f(y)\, {\it dy}\, {\it dx} =\left(\int_a^b f(x)\, {\it dx}\right)^2\). [HINT: Notice that \(\displaystyle\left(\int_a^b f(x)\, {\it dx}\right)^2=\intop\!\!\!\intop\nolimits_{[a,b]\times [a,b]}f(x)f(y)\, {\it dx}\, {\it dy}\).]
Show that (for reference see Section 13.4, twentyninth exercise) \begin{eqnarray*} \frac{d}{\it dx} \int_a^x\int_c^d f(x,y,z)\, {\it dz}\, {\it dy} &=&\int_c^d f(x,y,z)\, {\it dz}\\ &&+\ \int_a^x\int_c^d f_x(x,y,z)\, {\it dz}\, {\it dy} . \end{eqnarray*}