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On the disc $z=0$, the upward normal selects only the third component, which equals $e^0/(1+0)=1$. Hence the surface integral is the area of the unit disk:

The two orthogonal directions $(1,1{)}^T$ and $(1,-1{)}^T$ are preserved by this symmetric matrix. Direct multiplication, using the hyperbolic function identities $\cosh u\pm \sinh u=e^{\pm u}$, gives the eigenvalue–eigenvector pairsEach eigenvector can be multiplied by any nonzero scalar. For $u\ne 0$ these give the two one-dimensional eigenspaces. When $u=0$, the matrix is the identity: the only eigenvalue is $1$, with multiplicity two, and every nonzero vector is an eigenvector.

The tangent half-angle substitution gives $\sin \theta =2t/(1+t^2)$ and $\cos \theta =(1-t^2)/(1+t^2)$. Dividing and then taking the reciprocal square of the cosine givesThese formulas apply where $t$ is defined and $t\ne \pm 1$; at $t=\pm 1$ the original tangent and secant function are undefined because $\cos \theta =0$.

Use integration by substitution with $u=x^4$, so $du=4x^{3}dx$. ThenDifferentiating the answer by the chain rule gives $x^3{e}^{-x^4}$, confirming the antiderivative. The real constant of integration $C$ is arbitrary.

Let $p(x)=x^3-5x-1$. Its values at useful points areSince a polynomial is a continuous function, the intermediate value theorem gives one polynomial root in each of the disjoint intervals $(-3,-2)$, $(-1,0)$ and $(2,3)$. A nonzero cubic polynomial has at most three distinct polynomial roots. Hence it hasThese three distinct polynomial roots already exhaust its degree of a polynomial, so each has multiplicity of a root one.

Both segments start at $(1,1,1)$, so their direction vectors are $\mathbf{a}=(-1,-1,-1)$ and $\mathbf{b}=(1,1,0)$. Their dot product is $-2$, and their norms are $\sqrt{3}$ and $\sqrt{2}$. ThusThe angle between the rays lies in $[0,\pi ]$, so its sine is nonnegative:The negative cosine correctly identifies an obtuse angle. One also obtains the sine from $|\mathbf{a}\times \mathbf{b}|/(|\mathbf{a}||\mathbf{b}|)=\sqrt{2}/\sqrt{6}$, using the cross product.

For $f(x)=x{e}^{-2x^2}$, the product rule and chain rule giveThe positive exponential factor makes $f$ increasing for $0<x<1/2$ and decreasing for $1/2<x<1$. Its only interior stationary point is therefore a maximum at $x=1/2$. Checking it together with both endpoints,ConsequentlyChecking the endpoints is necessary for extrema on a closed interval.

Both sums are geometric series. For first term $a$ and common ratio $r$ with $|r|<1$, the sum is $a/(1-r)$. Taking the initial term to be $2$ givesandBoth have absolute convergence because the series of absolute values has ratio $1/2$.

Using the Pythagorean trigonometric identity and the double-angle formula,The required value $1/2$ therefore occurs when ${\sin }^2(2\theta )=1$. Within $0\le \theta \le \pi$, the only possibilities are $2\theta =\pi /2$ and $3\pi /2$. HenceEquivalently equality requires ${\sin }^2\theta ={\cos }^2\theta =1/2$.

Find two numbers with sum $-1$ and product $-2$: they are $-2$ and $1$. Thus the quadratic polynomial factorizes asThe product is negative precisely when its two factors have opposite signs. This occurs between its two polynomial roots, soThe endpoints give zero and are excluded by the strict inequality.