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

Ibelieve that c. recording would work best.

Answer from: Quest

For range i think it’s area

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Help, please this is calc 3
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