Mathematics, 19.12.2020 09:20 heathhm623

Help, please this is calc 3

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Answer from: PlaneGamer5678

$\int\limits^\pi_0 {\int\limits^{7cos(\theta)}_0 {r} \, dr } \, d\theta = \frac{49 \pi}{4}$

Top Right (Selected) Sketch

General Formulas and Concepts:

Calculus

Fundamental Theory of Calculus: $\int\limits^b_a {f(x)} \, dx = F(b) - F(a)$ Reverse Power Rule: $\int {x^n} \, dx = \frac{x^{n+1}}{n+1}$ Integration Property: $\int {cf(x)} \, dx = c\int {f(x)} \, dx$ Trig Reduction Formula: $\int\limits {cos^n(x)} \, dx = \frac{n-1}{n} \int {cos^{n-2}(x)} \, dx + \frac{cos^{n-1}(x)sin(x)}{n}$

Multivariable Calculus

Sketching the Region REvaluating Double Integrals

Step-by-step explanation:

Step 1: Define

$\int\limits^\pi_0 {\int\limits^{7cos(\theta)}_0 {r} \, dr } \, d\theta$

Step 2: Integrate Pt. 1

Evaluate the inner integral.

Define: $\int\limits^{7cos(\theta)}_0 {r} \, dr }$ Integrate [Reverse Power Rule/FTC]: $\int\limits^{7cos(\theta)}_0 {r} \, dr } = \frac{r^{1+1}}{1+1} | \limits^{7cos(\theta)}_0$ Simplify: $\int\limits^{7cos(\theta)}_0 {r} \, dr } = \frac{r^2}{2} | \limits^{7cos(\theta)}_0$ Evaluate [FTC]: $\int\limits^{7cos(\theta)}_0 {r} \, dr } = \frac{(7cos(\theta))^2}{2} - \frac{0^2}{2}$ Simplify: $\int\limits^{7cos(\theta)}_0 {r} \, dr } = \frac{49cos^2(\theta)}{2}$

Step 3: Integrate Pt. 2

Evaluate Double Integral.

[Inner Integral] Back-Substitute: $\int\limits^\pi_0 {\int\limits^{7cos(\theta)}_0 {r} \, dr } \, d\theta = \int\limits^\pi_0 {\frac{49cos^2(\theta)}{2} } \, d\theta$ [Integral] Integration Property: $\int\limits^\pi_0 {\int\limits^{7cos(\theta)}_0 {r} \, dr } \, d\theta = \frac{49}{2} \int\limits^\pi_0 {cos^2(\theta) } \, d\theta$ [Integral] Trig Reduction Formula: $\int {cos^2(\theta) } \, d\theta = \frac{2-1}{2} \int {cos^{2-2}}(\theta) \, d\theta + \frac{cos^{2-1}(\theta)sin(\theta)}{2}$ [TRF Integral] Simplify: $\int {cos^2(\theta) } \, d\theta = \frac{1}{2} \int {1 \, d\theta + \frac{cos(\theta)sin(\theta)}{2}$ [TRF Integral] Integrate: $\int {cos^2(\theta) } \, d\theta = \frac{1}{2}(\theta) + \frac{cos(\theta)sin(\theta)}{2}$ [TRF Integral]: Simplify: $\int {cos^2(\theta) } \, d\theta = \frac{cos(\theta)sin(\theta)}{2} + \frac{\theta}{2}$ [Double Integral] Back-Substitute/FTC: $\int\limits^\pi_0 {\int\limits^{7cos(\theta)}_0 {r} \, dr } \, d\theta = \frac{49}{2} (\frac{cos(\theta)sin(\theta)}{2} + \frac{\theta}{2} ) | \limits^\pi_0$ [Double Integral] Evaluate FTC: $\int\limits^\pi_0 {\int\limits^{7cos(\theta)}_0 {r} \, dr } \, d\theta = \frac{49}{2} [(\frac{cos(\pi)sin(\pi)}{2} + \frac{\pi}{2} ) - (\frac{cos(0)sin(0)}{2} + \frac{0}{2} )]$ [Double Integral] Evaluate/Simplify: $\int\limits^\pi_0 {\int\limits^{7cos(\theta)}_0 {r} \, dr } \, d\theta = \frac{49}{2} [\frac{\pi}{2} - 0]$ [Double Integral] Evaluate: $\int\limits^\pi_0 {\int\limits^{7cos(\theta)}_0 {r} \, dr } \, d\theta = \frac{49 \pi}{4}$

Sketching out the Region R for this, we use our polar coordinate plane. Our angle Θ is 0 and our radius r is 0.

We look at the limits of integration for the radius r. Our r has to be between 0 and 7cos(Θ). The lowest possible value that can be attained by radius r is 0. The region has to be outside the circle r = 0. The largest possible value that can be attained by radius r is 7cos(Θ). Our region has to lie in this shape.

We now look at the limits of integration for the angle Θ. Our Θ has to be between 0 and π. Our region has to lie in that angle.

The top right graph (the one selected) is the correct sketch of the shaded region R.

Attached image is how we graph the region. Special thanks to ASimpleEngineer:

Answer from: Quest

Ac equals 35 cm long
If line ab = 21 centimeters, and line bc = 28 centimeters, which does line ac equal?

Answer from: Quest

step-by-step explanation:

use context clues to figure out the meaning of the bolded word:

the sage gardener knew when she had to water her plants.

question 8 options:

mint plant

wise

happy

lazy

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Help, please this is calc 3
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