Problem 60 - Calculus Early Transcendentals

Question

Two integrals Let \(R=\\{(x, y): 0 \leq x \leq 1,0 \leq y \leq 1\\}\) a. Evaluate \(\iint_{R} \cos (x \sqrt{y}) d A\) b. Evaluate \(\iint_{R} x^{3} y \cos \left(x^{2} y^{2}\right) d A\)

Step-by-Step Solution

Answer

Question: Evaluate the following integrals over the region R which is defined as a rectangle in the xy-plane with vertices (0,0), (1,0), (0,1), and (1,1): a. \(\iint_{R} \cos (x \sqrt{y}) d A\) b. \(\iint_{R} x^{3} y \cos \left(x^{2} y^{2}\right) d A\) Answer: a. \(\iint_{R} \cos (x \sqrt{y}) d A = -2\cos(1) + 2\) b. \(\iint_{R} x^{3} y \cos \left(x^{2} y^{2}\right) d A = \frac{1}{2}(1 - \cos(1))\)

Step 1: Set up the iterated integral

Since we are given the rectangular region R, we can directly write the iterated integral as: $$\int_{0}^{1} \int_{0}^{1} \cos (x \sqrt{y})\, dy\,dx$$

Step 2: Evaluate the inner integral

Consider the inner integral first: $$\int_{0}^{1} \cos (x \sqrt{y})\, dy$$ Let \(u = x\sqrt{y}\) which implies \(du = \frac{x}{2\sqrt{y}}\, dy\). Solving for dy, we get \(dy = \frac{2\,du}{x}\). Then our integral becomes: $$\int \cos(u)\, \frac{2\,du}{x}$$ Now, we substitute the bounds of integration for y in terms of u. When \(y = 0\), \(u = x\sqrt{0} = 0\), and when \(y = 1\), \(u = x\sqrt{1} = x\). Therefore, our integral becomes: $$\frac{2}{x}\int_{0}^{x} \cos(u)\, du$$

Step 3: Solve the inner integral

Now we can solve the inner integral: $$\frac{2}{x}\int_{0}^{x} \cos(u)\, du = \frac{2}{x} [\sin(u)]_{0}^{x} = \frac{2}{x}(\sin(x) - \sin(0)) = 2\sin(x)$$

Step 4: Evaluate the outer integral

Now evaluate the outer integral: $$\int_{0}^{1} 2\sin(x)\, dx$$ Solve the integral: $$[ -2\cos(x)]_{0}^{1} = -2\cos(1) + 2\cos(0) = -2\cos(1) + 2$$ So, the value of the integral \(\iint_{R} \cos (x \sqrt{y}) d A = -2\cos(1) + 2\). b. Evaluate \(\iint_{R} x^{3} y \cos \left(x^{2} y^{2}\right) d A\)

Step 1: Set up the iterated integral

Similar to part a, we can directly write the iterated integral as: $$\int_{0}^{1} \int_{0}^{1} x^{3} y \cos \left(x^{2} y^{2}\right)\, dy\,dx$$

Step 2: Evaluate the inner integral

Consider the inner integral: $$\int_{0}^{1} x^{3} y \cos \left(x^{2} y^{2}\right)\, dy$$ Let \(u = x^2 y^2\). Then, \(du = 2x^2 y\, dy\). Solving for dy, we get \(dy = \frac{du}{2x^2 y}\). Now, our integral becomes: $$x^2\,\int \cos(u)\, \frac{du}{2}$$ Substitute the bounds of integration for y in terms of u. When \(y = 0\), \(u = x^2\cdot0^2 = 0\), and when \(y = 1\), \(u = x^2\cdot1^2 = x^2\). Therefore, our integral becomes: $$\frac{x^2}{2}\int_{0}^{x^2} \cos(u)\, du$$

Step 3: Solve the inner integral

Now we can solve the inner integral: $$\frac{x^2}{2}\int_{0}^{x^2} \cos(u)\, du = \frac{x^2}{2} [\sin(u)]_{0}^{x^2} = \frac{x^2}{2} (\sin(x^2)-\sin(0)) = \frac{x^2}{2}\sin(x^2)$$

Step 4: Evaluate the outer integral

Now evaluate the outer integral: $$\int_{0}^{1} \frac{x^2}{2}\sin(x^2)\, dx$$ Let \(v = x^2\) and \(dv = 2x\, dx\). Then our integral becomes $$\frac{1}{2}\int_0^1 v\sin{(v)}\,dv$$ Now, use integration by parts: Let $p = v \Rightarrow dp = dv \\ q' = \sin(v) \Rightarrow q = -\cos(v)$ Now, integration by parts formula \(pq - \int{qdp}\) \(-\frac{1}{2}(-v\cos{(v)}-\int{-\cos{(v)}dv)_{0}^{1}}=\frac{1}{2}(1 - \cos(1))\) So, the value of the integral \(\iint_{R} x^{3} y \cos \left(x^{2} y^{2}\right) d A = \frac{1}{2}(1 - \cos(1))\).

Key Concepts

Iterated Integrals

When evaluating double integrals over a region, especially a rectangular one like the region \(R = \{(x, y): 0 \leq x \leq 1,0 \leq y \leq 1\}\), we often use a process called iterated integrals. This involves breaking down a double integral into an integral of integrals. Simply put, an iterated integral is a double integral solved as two single integrals performed sequentially. The order you solve them—whether you integrate with respect to \(y\) first or \(x\)—depends on the problem and limits but often follows region geometry first. For example, in our exercise, the iterated integral is set up as:

• First, integrate \( \int_{0}^{1} \int_{0}^{1} \cos (x \sqrt{y}) \, dy\,dx \) with respect to \(y\), keeping \(x\) constant, and then
• Integrate the result with respect to \(x\).

The key is to pay careful attention to the bounds as they will change according to the variable you integrate first. Understanding this process helps solve complex integrals simply by repeating familiar integration techniques sequentially.

Integration by Parts

Integration by parts is a powerful technique that allows us to transform complex integrals into simpler ones. It's based on the product rule for differentiation and helps in solving integrals where a simple standard method doesn’t suffice.The formula is: \[ \int u \, dv = uv - \int v \, du \]You select parts of the integrand as \(u\) and \(dv\). For our case, when evaluating the outer integral:

• We took \(v = x^2\) resulting in \(dv = 2x \, dx\).
• Also, we used \(q' = \sin(v)\), which integrates to \(q = -\cos(v)\).

This let us transform \( \int v \sin(v) \, dv \) into simpler terms \( v(-\cos(v)) - \int (-\cos(v))\, dv \). The process simplifies the problem by reducing the integral into more manageable components. Hence, with practice, integration by parts becomes more intuitive and a valuable tool in your integration toolkit.

Trigonometric Integrals

Trigonometric integrals consist of integrals that involve trigonometric functions like \(\sin\), \(\cos\), and others. They often appear in calculus problems and require specific strategies to solve.In the exercise, the trigonometric function \(\cos\) appears as \(\cos(x \sqrt{y})\) within the integral. Solving such integrals might need substituting variables smartly to align them with known integral identities or performing integration by parts.For example, with the integral \( \int \cos(u) \, du \), we directly apply the identity \( \int \cos(u) \, du = \sin(u) + C \). It’s also common to see trigonometric functions in repeated patterns due to their periodic nature, which further simplifies through strategic substitutions. Understanding these basic trigonometric integrals and their properties is crucial since they serve as building blocks for handling more complex integrals.