Math Problem Statement

Evaluate the iterated integral by converting to polar coordinates.

NOTE: When typing your answers use "th " for θ . ∫6√0∫12−y2√y2x+3y dx dy=∫ba∫dc

functionsequation editor dr dθ

where

a=

0 functionsequation editor b=

pi/4 functionsequation editor c=

0 functionsequation editor d=

3.464 functionsequation editor ∫6√0∫12−y2√y2x+3y dx dy=

Solution

To evaluate the given iterated integral:

∫06∫012−y2(y2x+3y) dx dy,\int_0^{\sqrt{6}} \int_0^{\sqrt{12-y^2}} (y^2x + 3y) \, dx \, dy,

we will convert it to polar coordinates. Recall the polar coordinate transformations: x=rcos⁡θ,y=rsin⁡θ,anddx dy=r dr dθ.x = r\cos\theta, \quad y = r\sin\theta, \quad \text{and} \quad dx\,dy = r\,dr\,d\theta.

Step 1: Interpret the bounds

  • The region of integration is bounded by 0≤y≤60 \leq y \leq \sqrt{6} and 0≤x≤12−y20 \leq x \leq \sqrt{12 - y^2}.
  • These bounds describe a quarter-circle in the first quadrant with a radius of 12\sqrt{12}.

Thus, the polar bounds are: 0≤r≤12,0≤θ≤π4.0 \leq r \leq \sqrt{12}, \quad 0 \leq \theta \leq \frac{\pi}{4}.

Step 2: Rewrite the function in polar coordinates

The function y2x+3yy^2x + 3y becomes:

= r^3\sin^2\theta\cos\theta + 3r\sin\theta.$$ ### Step 3: Convert the integral The iterated integral becomes: $$\int_0^{\pi/4} \int_0^{\sqrt{12}} \left(r^3\sin^2\theta\cos\theta + 3r\sin\theta\right) r \, dr \, d\theta.$$ Expanding $$r \cdot r^3$$ gives $$r^4$$, and $$r \cdot 3r$$ gives $$3r^2$$: $$\int_0^{\pi/4} \int_0^{\sqrt{12}} \left(r^4\sin^2\theta\cos\theta + 3r^2\sin\theta\right) \, dr \, d\theta.$$ ### Step 4: Separate the integral We can split this into two integrals: $$\int_0^{\pi/4} \int_0^{\sqrt{12}} r^4\sin^2\theta\cos\theta \, dr \, d\theta + \int_0^{\pi/4} \int_0^{\sqrt{12}} 3r^2\sin\theta \, dr \, d\theta.$$ ### Step 5: Solve the $$r$$-integrals #### First term: $$\int_0^{\sqrt{12}} r^4 \, dr = \left[\frac{r^5}{5}\right]_0^{\sqrt{12}} = \frac{(\sqrt{12})^5}{5} = \frac{12\sqrt{12}}{5}.$$ This makes the first term: $$\frac{12\sqrt{12}}{5} \int_0^{\pi/4} \sin^2\theta\cos\theta \, d\theta.$$ #### Second term: \[ \int_0^{\sqrt{12}} 3r^2 \, dr = 3 \left[\frac{r^3}{3}\right]_0^{\sqrt{12}} = \left[(\sqrt{12})^3\right] = 12\sqrt{12}. \] This makes the second term: $$12\sqrt{12} \int_0^{\pi/4} \sin\theta \, d\theta.$$ ### Step 6: Solve the $$\theta$$-integrals #### For $$\int_0^{\pi/4} \sin^2\theta\cos\theta \, d\theta$$: Use the substitution $$u = \sin\theta$$, so $$du = \cos\theta \, d\theta$$. When $$\theta = 0$$, $$u = 0$$; when $$\theta = \pi/4$$, $$u = \sin(\pi/4) = \frac{\sqrt{2}}{2}$$: $$\int_0^{\pi/4} \sin^2\theta\cos\theta \, d\theta = \int_0^{\sqrt{2}/2} u^2 \, du = \left[\frac{u^3}{3}\right]_0^{\sqrt{2}/2} = \frac{(\sqrt{2}/2)^3}{3} = \frac{\sqrt{2}}{24}.$$ #### For $$\int_0^{\pi/4} \sin\theta \, d\theta$$: $$\int_0^{\pi/4} \sin\theta \, d\theta = \left[-\cos\theta\right]_0^{\pi/4} = -\cos\left(\frac{\pi}{4}\right) + \cos(0) = -\frac{\sqrt{2}}{2} + 1.$$ ### Step 7: Combine the results Substituting back: - First term: $$\frac{12\sqrt{12}}{5} \cdot \frac{\sqrt{2}}{24} = \frac{\sqrt{12} \cdot \sqrt{2}}{10} = \frac{\sqrt{24}}{10} = \frac{2\sqrt{6}}{10} = \frac{\sqrt{6}}{5}.$$ - Second term: $$12\sqrt{12} \cdot \left(1 - \frac{\sqrt{2}}{2}\right) = 12\sqrt{12} - 6\sqrt{24}.$$ Thus, the value of the integral is: $$\frac{\sqrt{6}}{5} + 12\sqrt{12} - 6\sqrt{24}.$$

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Polar Coordinates
Multivariable Calculus
Integration

Formulas

x = r cos(θ), y = r sin(θ), dx dy = r dr dθ
Integral bounds for polar coordinates: r and θ
Trigonometric Substitution for Integration

Theorems

Change of Variables Theorem
Trigonometric Identities

Suitable Grade Level

Undergraduate Calculus (Grades 11-12 or College)

Related Recommendation

Double Integral of x^2 y^2 over a Circular Region Using Polar Coordinates

Evaluate Integral from 0 to 4 with Change of Variables and Substitution

Evaluate the Double Integral \( \int_0^2 \int_0^{\sqrt{x}} x^3 y \, dy \, dx \)

Evaluate Double Integral Using Polar Coordinates

Solve Double Integral by Changing to Polar Coordinates