\$intint_D x^2y , dA\$, wright here \$D\$ is the height half of the disk via center the beginning and also radius \$5\$.

You are watching: Evaluate the given integral by changing to polar coordinates

as soon as i plugged every little thing in, I obtained the double integral \$\$int_0^pi int_0^5 r^4cos^2 heta sin heta ,dr ,d heta.\$\$ then I provided \$u\$ substitution and also gained \$5int u^2 ,du =left. frac5u^33 ightvert_ heta=0^ heta=pi =-5/3 - 5/3 = -10/3\$ yet that"s definitely wrong. I looked at multiple examples, what am I doing wrong here? I"m pretty sure i have actually the appropriate method however something"s going wrong.

You have forobtained to integrate via respect to \$r\$ and also I have actually no idea wbelow does your leading \$5\$ comes from.

\$\$int_0^5r^4 ,dr =frac5^55=5^4\$\$

eginalignint_0^pi cos^2 heta sin heta , d heta &= left. -fraccos^3 heta3 ightvert_0^piendalign

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Evaluate the provided integral by altering to polar coordinates. (exactly how to combine \$sin(r^2)r\$ through respect to \$r\$?

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