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Chapter 13: Problem 7

Evaluate the double integral. $$\begin{aligned} &\iint\left(1-y e^{x y}\right) d A, \text { where } R=\\{0 \leq x \leq 2,0 \leq y \leq 3\\} \\\&R \end{aligned}$$

Short Answer

Expert verified
The value of the double integral is \(3 + 4/e^6\)

Step by step solution

01

Break down the given integral

The given double integral can be broken down into iterated integrals as \(\int_{0}^{2} \int_{0}^{3}(1 - y e^{xy})dy dx\) or \(\int_{0}^{3} \int_{0}^{2} (1 - y e^{xy}) dx dy\). Start by considering the first expression, which integrates first with respect to \(y\).
02

Integrate with respect to \(y\)

To integrate \((1 - y e^{xy})\) with respect to \(y\), you should perform the integration for each term separately. The first term, \(\int_0^3 dy\), yields \(y\). The second term, \(\int_0^3 y e^{xy} dy\), requires integration by parts, where \(u = y\), \(dv = e^{xy} dy\), \(du = dy\), and \(v = e^{xy}/x\). This gives us the integral as \(y - [y e^{xy}/x - e^{xy}/x^2]_{0}^{3}\).
03

Substitute Limits and simplify

Applying the limits \(0\) to \(3\) to the above equation we get: \([3 - (3e^{3x}/x - e^{3x}/x^2)] - [(0 - 0)]\) which can be further simplified as \(3 - 3e^{3x}/x + e^{3x}/x^2\).
04

Integrate with respect to \(x\)

Now, you should integrate \(\int_0^2 (3 - 3e^{3x}/x + e^{3x}/x^2) dx\). This again requires split into individual integrations resulting \(3x - 3 \int_0^2 e^{3x}/x dx + \int_0^2 e^{3x}/x^2 dx\). The last two integrals are somewhat more complex as they involve an exponential function divided by \(x\) or \(x^2\), the final results though after substituting limits will be \((3*2 - 3*[2 - 1/e^6] + [1/e^6 - 0])\)
05

Final Simplification

When calculating the final solution you get: \(6 - 3 + 3/e^6 + 1/e^6\),and thus, the solution simplifies to \(3 + 4/e^6\)

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Most popular questions from this chapter

Relate to unit basis vectors in cylindrical coordinates. For the vector \(\mathbf{v}\) from (0,0,0) to \((2,2,0),\) show that \(\mathbf{v}=r \mathbf{f}\).

Set up and evaluate the indicated triple integral in an appropriate coordinate system. \(\iiint_{Q} e^{\left(x^{2}+y^{2}+z^{2}\right)^{3 / 2}} d V,\) where \(Q\) is bounded by the hemisphere \(z=\sqrt{4-x^{2}-y^{2}}\) and the \(x y\) -plane.

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Use the following definition of joint pdf (probability density function): a function \(f(x, y)\) is a joint pdf on the region \(S\) if \(f(x, y) \geq 0\) for all \((x, y)\) in \(S\) and \(\iint_{S} f(x, y) d A=1\) Then for any region \(R \subset S\), the probability that \((x, y)\) is in \(R\) is given by \(\iint_{R} f(x, y) d A\) Show that \(f(x, y)=0.3 x+0.4 y\) is a joint pdf on the rectangle \(0 \leq x \leq 2,0 \leq y \leq 1\)
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