document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1210245>Question 1210245</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Evaluate the double integral by converting it into polar coordinates:  integral from 1 to 2 integral from 0 to sqrt (2x - x^2) (x^2y + y^3) dy dx\r<BR>\n" );
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document.write( "I can get the answer using rectangular coords as 47/60, but conversion into polar and evaluation gets me the wrong answer. I can't seem to get the correct limits or whatever. <BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #851775 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1959)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=851775\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Certainly! Let's evaluate this double integral by converting it to polar coordinates.\r<BR>\n" );
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document.write( "First, we need to understand the region of integration defined by the limits:<BR>\n" );
document.write( "$$1 \le x \le 2$$<BR>\n" );
document.write( "$$0 \le y \le \sqrt{2x - x^2}$$\r<BR>\n" );
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document.write( "The upper limit for $y$ can be rewritten as $y^2 = 2x - x^2$, which leads to $x^2 - 2x + y^2 = 0$. Completing the square for the $x$ terms, we get $(x - 1)^2 + y^2 = 1$. This equation represents a circle with center $(1, 0)$ and radius 1.\r<BR>\n" );
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document.write( "The condition $y \ge 0$ restricts us to the upper half of this circle. The limits on $x$, $1 \le x \le 2$, further confine the region to the part of the upper semicircle where $x$ ranges from 1 to 2.\r<BR>\n" );
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document.write( "Now, let's convert to polar coordinates using the transformations:<BR>\n" );
document.write( "$$x = r \cos \theta$$$$y = r \sin \theta$$$$dy \, dx = r \, dr \, d\theta$$\r<BR>\n" );
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document.write( "The equation of the circle $(x - 1)^2 + y^2 = 1$ in polar coordinates becomes:<BR>\n" );
document.write( "$$(r \cos \theta - 1)^2 + (r \sin \theta)^2 = 1$$$$r^2 \cos^2 \theta - 2r \cos \theta + 1 + r^2 \sin^2 \theta = 1$$$$r^2 (\cos^2 \theta + \sin^2 \theta) - 2r \cos \theta = 0$$$$r^2 - 2r \cos \theta = 0$$$$r(r - 2 \cos \theta) = 0$$<BR>\n" );
document.write( "Since $r \ge 0$, we have $r = 2 \cos \theta$.\r<BR>\n" );
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document.write( "To determine the limits for $\theta$, we observe the region of integration. The circle intersects the x-axis at $x=0$ and $x=2$. The part of the circle we are interested in corresponds to angles from $\theta = 0$ (where $x=r$) up to where the circle intersects the y-axis tangentially at $x=1$, which corresponds to $\theta = \frac{\pi}{2}$.\r<BR>\n" );
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document.write( "The limits for $r$ will go from $0$ to the boundary of the region, which is given by $r = 2 \cos \theta$.\r<BR>\n" );
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document.write( "The integrand $x^2y + y^3$ can be written in polar coordinates as:<BR>\n" );
document.write( "$$(r \cos \theta)^2 (r \sin \theta) + (r \sin \theta)^3 = r^3 \cos^2 \theta \sin \theta + r^3 \sin^3 \theta = r^3 \sin \theta (\cos^2 \theta + \sin^2 \theta) = r^3 \sin \theta$$\r<BR>\n" );
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document.write( "Now we can set up the double integral in polar coordinates:<BR>\n" );
document.write( "$$\int_{0}^{\pi/2} \int_{0}^{2 \cos \theta} (r^3 \sin \theta) \cdot r \, dr \, d\theta = \int_{0}^{\pi/2} \int_{0}^{2 \cos \theta} r^4 \sin \theta \, dr \, d\theta$$\r<BR>\n" );
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document.write( "Let's evaluate the inner integral with respect to $r$:<BR>\n" );
document.write( "$$\int_{0}^{2 \cos \theta} r^4 \sin \theta \, dr = \sin \theta \left[ \frac{r^5}{5} \right]_{0}^{2 \cos \theta} = \sin \theta \left( \frac{(2 \cos \theta)^5}{5} - 0 \right) = \frac{32}{5} \cos^5 \theta \sin \theta$$\r<BR>\n" );
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document.write( "Now, we evaluate the outer integral with respect to $\theta$:<BR>\n" );
document.write( "$$\int_{0}^{\pi/2} \frac{32}{5} \cos^5 \theta \sin \theta \, d\theta$$Let $u = \cos \theta$, then $du = -\sin \theta \, d\theta$. When $\theta = 0$, $u = 1$, and when $\theta = \pi/2$, $u = 0$.$$\frac{32}{5} \int_{1}^{0} u^5 (-du) = \frac{32}{5} \int_{0}^{1} u^5 \, du = \frac{32}{5} \left[ \frac{u^6}{6} \right]_{0}^{1} = \frac{32}{5} \left( \frac{1^6}{6} - \frac{0^6}{6} \right) = \frac{32}{5} \cdot \frac{1}{6} = \frac{32}{30} = \frac{16}{15}$$\r<BR>\n" );
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document.write( "Thus, the value of the double integral is $\frac{16}{15}$.\r<BR>\n" );
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document.write( "Final Answer: The final answer is $\boxed{\frac{16}{15}}$ </SPAN>\n" );
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