Read the Original Region
For
\[\int_a^b\int_{g_1(x)}^{g_2(x)}f(x,y)\,dy\,dx,\]
the vertical strip runs from \(y=g_1(x)\) to \(y=g_2(x)\).
Changing the Order of Integration
Sketching the Region and Reversing the Limits
Changing the order of integration is developed through three IOE past-paper questions. Keep the shaded region unchanged, reverse the direction of the strip and write the new limits carefully.
UnitMultiple Integrals
TopicChange of Order
Question Type4 Mark Problems
Theory Required for These Problems
Reverse the Order
Sketch the same shaded region and describe it with a horizontal strip:
\[h_1(y)\le x\le h_2(y).\]
Then write the new outside limits for \(y\).
Split When Required
If one horizontal boundary changes at an intersection or turning point, split the integral into two parts. Both parts must cover the original region without overlap.
Final Check
Check the corner points, the direction of the strip and the order of the differentials. The inner limits must match the inner differential.
Note: The shaded region remains unchanged. Only the direction of the strip changes.
Method used: Draw the region, mark the intersections, reverse the strip, rewrite the limits, split the region when necessary, and then integrate.
Continue through the IOE Engineering Mathematics II Digital Workbook. This changing the order of integration page follows the Engineering Mathematics II course sequence in the Tribhuvan University curriculum.
13
4 Marks
Infinite Integral by Changing the Order
PAST-PAPER REFERENCE: TU IOE • 2083 Baisakh • Back (New Course) • ENSH 151 • Q. 8 • 4 Marks
Evaluate the following integral by changing the order of integration:
\[\int_0^{\infty}\int_x^{\infty}\frac{e^{-xy}}{y}\,dy\,dx.\]
\[\int_0^{\infty}\int_x^{\infty}\frac{e^{-xy}}{y}\,dy\,dx.\]
Solution
Change of order
The original limits give
\[x\in[0,\infty),\qquad y\in[x,\infty).\]After changing the order,
\[y\in[0,\infty),\qquad x\in[0,y].\]Therefore,
\[I=\int_0^{\infty}\int_0^y\frac{e^{-xy}}{y}\,dx\,dy.\]Integrating first with respect to x,
\[I=\int_0^{\infty}\frac{1-e^{-y^2}}{y^2}\,dy.\]Integrate by parts. Take
\[u=1-e^{-y^2},\qquad dv=y^{-2}\,dy.\]Then
\[du=2ye^{-y^2}\,dy,\qquad v=-\frac1y.\]Hence,
\[I=\left[-\frac{1-e^{-y^2}}{y}\right]_0^{\infty}+2\int_0^{\infty}e^{-y^2}\,dy.\]
\[I=\left[-\frac{1-e^{-y^2}}{y}\right]_0^{\infty}\]\[{}+2\int_0^{\infty}e^{-y^2}\,dy.\]
The boundary term is zero and
\[\int_0^{\infty}e^{-y^2}\,dy=\frac{\sqrt\pi}{2}.\]Therefore,
\[I=2\left(\frac{\sqrt\pi}{2}\right)=\sqrt\pi.\]\[I=\sqrt\pi\]
14
4 Marks
Region Split after Changing the Order
PAST-PAPER REFERENCE: TU IOE • 2082 Bhadra • Regular (New Course) • ENSH 151 • Q. 8 • 4 Marks
Change the order of integration and evaluate
\[\int_0^1\int_{x^2}^{2-x}xy\,dy\,dx.\]
\[\int_0^1\int_{x^2}^{2-x}xy\,dy\,dx.\]
Solution
Split the horizontal description
The curves are
\[y=x^2\qquad\mathrm{and}\qquad y=2-x.\]For the lower part,
\[0\le y\le1,\qquad0\le x\le\sqrt y.\]For the upper part,
\[1\le y\le2,\qquad0\le x\le2-y.\]Therefore,
\[I=\int_0^1\int_0^{\sqrt y}xy\,dx\,dy+\int_1^2\int_0^{2-y}xy\,dx\,dy.\]
\[I=\int_0^1\int_0^{\sqrt y}xy\,dx\,dy\]\[{}+\int_1^2\int_0^{2-y}xy\,dx\,dy.\]
Integrating with respect to x,
\[I=\frac12\int_0^1y^2\,dy+\frac12\int_1^2y(2-y)^2\,dy.\]
\[I=\frac12\int_0^1y^2\,dy\]\[{}+\frac12\int_1^2y(2-y)^2\,dy.\]
Now,
\[\frac12\int_0^1y^2\,dy=\frac16.\]Also,
\[\frac12\int_1^2y(2-y)^2\,dy=\frac5{24}.\]Hence,
\[I=\frac16+\frac5{24}=\frac38.\]\[I=\frac38\]
15
4 Marks
Triangular Region after Changing the Order
PAST-PAPER REFERENCE: TU IOE • 2081 Ashwin • Regular (New Course, 2080 Batch) • SH 151 • Q. 8 • 4 Marks
Change the order of integration and evaluate
\[\int_0^a\int_0^x\frac{\cos y}{\sqrt{(a-x)(a-y)}}\,dy\,dx.\]
\[\int_0^a\int_0^x\frac{\cos y}{\sqrt{(a-x)(a-y)}}\,dy\,dx.\]
Solution
Change of order in a triangular region
The original region is
\[0\le x\le a,\qquad0\le y\le x.\]After changing the order,
\[0\le y\le a,\qquad y\le x\le a.\]Therefore,
\[I=\int_0^a\int_y^a\frac{\cos y}{\sqrt{(a-x)(a-y)}}\,dx\,dy.\]Take the factors independent of x outside:
\[I=\int_0^a\frac{\cos y}{\sqrt{a-y}}\left[\int_y^a\frac{dx}{\sqrt{a-x}}\right]dy.\]
\[I=\int_0^a\frac{\cos y}{\sqrt{a-y}}\]\[\left[\int_y^a\frac{dx}{\sqrt{a-x}}\right]dy.\]
Now,
\[\int_y^a\frac{dx}{\sqrt{a-x}}=\left[-2\sqrt{a-x}\right]_y^a=2\sqrt{a-y}.\]
\[\int_y^a\frac{dx}{\sqrt{a-x}}\]\[=\left[-2\sqrt{a-x}\right]_y^a\]\[=2\sqrt{a-y}.\]
Hence,
\[I=2\int_0^a\cos y\,dy.\]Therefore,
\[I=2[\sin y]_0^a=2\sin a.\]\[I=2\sin a\]