Add and Subtract Rational Expressions with a Common Denominator

Mathematics

1 answer:

Sladkaya [172]2 years ago

7 0

Answer:

$\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}$

Step-by-step explanation:

<u>Simplifying Rational Expressions</u>

If two or more rational expressions have the same denominator, the add and subtract operations are done only with the numerator. The final denominator will be the common of both.

The expression is:

$\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}$

Operating on the numerators:

$\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-(3q^2-q-6)}{q^2+6q+5}$

Removing parentheses:

$\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-3q^2+q+6}{q^2+6q+5}$

Simplifying:

$\boxed{\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}}$

The expression cannot be further simplified.

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Rewrite the following integral in spherical coordinates.

lora16 [44]

In cylindrical coordinates, we have $r^2=x^2+y^2$ , so that

$z = \pm \sqrt{2-r^2} = \pm \sqrt{2-x^2-y^2}$

correspond to the upper and lower halves of a sphere with radius $\sqrt2$ . In spherical coordinates, this sphere is $\rho=\sqrt2$ .

$1 \le r \le \sqrt2$ means our region is between two cylinders with radius 1 and $\sqrt2$ . In spherical coordinates, the inner cylinder has equation

$x^2+y^2 = 1 \implies \rho^2\cos^2(\theta) \sin^2(\phi) + \rho^2\sin^2(\theta) \sin^2(\phi) = \rho^2 \sin^2(\phi) = 1 \\\\ \implies \rho^2 = \csc^2(\phi) \\\\ \implies \rho = \csc(\phi)$

This cylinder meets the sphere when

$x^2 + y^2 + z^2 = 1 + z^2 = 2 \implies z^2 = 1 \\\\ \implies \rho^2 \cos^2(\phi) = 1 \\\\ \implies \rho^2 = \sec^2(\phi) \\\\ \implies \rho = \sec(\phi)$

which occurs at

$\csc(\phi) = \sec(\phi) \implies \tan(\phi) = 1 \implies \phi = \dfrac\pi4+n\pi$

where $n\in\Bbb Z$ . Then $\frac\pi4\le\phi\le\frac{3\pi}4$ .

The volume element transforms to

$dx\,dy\,dz = r\,dr\,d\theta\,dz = \rho^2 \sin(\phi) \, d\rho \, d\theta \, d\phi$

Putting everything together, we have

$\displaystyle \int_0^{2\pi} \int_1^{\sqrt2} \int_{-\sqrt{2-r^2}}^{\sqrt{2-r^2}} r \, dz \, dr \, d\theta = \boxed{\int_0^{2\pi} \int_{\pi/4}^{3\pi/4} \int_{\csc(\phi)}^{\sqrt2} \rho^2 \sin(\phi) \, d\rho \, d\phi \, d\theta} = \frac{4\pi}3$