Iterated Integrals — Question 2

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Question 2

Evaluate ∬Re2xcos⁡ydA,R=[0,ln⁡3]×[0,π3].\iint_R e^{2x}\cos y\,dA, \qquad R=[0,\ln 3]\times\left[0,\frac{\pi}{3}\right].

Tasks

  1. Recognize and use the product structure of the integrand.

  2. Evaluate the integral in each order.

  3. Verify the exact result by multiplying the two one-variable integrals.

Original worksheet page 1: question and worked solution for 4-2-002
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Question 2 – Solution

Strategy. The integrand is a product of an xx-only factor and a yy-only factor, so constant rectangular bounds make the two integrations separate.

Step 1: Integrate in yy first ∫0ln⁡3∫0π/3e2xcos⁡ydydx=∫0ln⁡3e2x[sin⁡y]0π/3dx=32[12e2x]0ln⁡3=34(9−1)=23.\begin{align*} \int_0^{\ln 3}\int_0^{\pi/3}e^{2x}\cos y\,dy\,dx &=\int_0^{\ln 3}e^{2x}[\sin y]_0^{\pi/3}\,dx\\ &=\frac{\sqrt 3}{2}\left[\frac 12e^{2x}\right]_0^{\ln 3}\\ &=\frac{\sqrt 3}{4}(9-1)=2\sqrt 3. \end{align*}

Step 2: Reverse the order ∫0π/3∫0ln⁡3e2xcos⁡ydxdy=∫0π/3cos⁡y[12e2x]0ln⁡3dy=4[sin⁡y]0π/3=23.\begin{align*} \int_0^{\pi/3}\int_0^{\ln 3}e^{2x}\cos y\,dx\,dy &=\int_0^{\pi/3}\cos y\left[\frac 12e^{2x}\right]_0^{\ln 3}dy\\ &=4[\sin y]_0^{\pi/3}=2\sqrt 3. \end{align*} Therefore ∬Re2xcos⁡ydA=23.\boxed{\iint_R e^{2x}\cos y\,dA=2\sqrt 3}.

Verification Direct factorization gives (∫0ln⁡3e2xdx)(∫0π/3cosydy)=4⋅32=23,\left(\int_0^{\ln 3}e^{2x}\,dx\right) \left(\int_0^{\pi/3}\cos y\,dy\right) =4\cdot\frac{\sqrt 3}{2}=2\sqrt 3, confirming both orders and the endpoint e2ln⁡3=9e^{2\ln 3}=9.

Original worksheet page 2: question and worked solution for 4-2-002

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