<p>7th Edition Barron’s (if you have a different edition it doesn’t matter, most of the questions remain the same so you should still be able to find it), Chapter 6: Definite Integrals</p>
<li>If x=4cosΘ and y=3sinΘ, then ∫xy dx (upper limit:4, lower limit 2) is equivalent to?</li>
</ol>
<p>A) 48∫sinΘ<a href=“UL:%200,%20LL:%20pi/3”>cosΘ^2</a>
B) 48∫[sinΘ^2]cosΘ (UL: 4, LL: 2)
C) 36∫(sinΘ)<a href=“UL:%204,%20LL:%202”>cosΘ^2</a>
D) -48∫(sinΘ)<a href=“UL:%20pi/3,%20LL:%200”>cosΘ^2</a>
E) 48∫[sinΘ^2]cosΘ (UL: pi/3, LL: 0)</p>
<p>Answer given is E</p>
<p>Could someone please explain this question in detail to me? I completely do not understand the explanation given by Barron’s.</p>
<p>Thanks :)</p>
<p>Basically what you are doing here is converting from (x,y) to Θ, so the first thing you have to do is change the bounds in order to make them into terms of Θ</p>
<p>Lower bound:
x = 4 cos Θ
2 = 4 cos Θ
0.5 = cos Θ
Θ = pi/3.</p>
<p>Upper bound:
x = 4 cos Θ
4 = 4 cos Θ
1 = cos Θ
Θ = 0</p>
<p>Now you have to get dx in terms of Θ.
x = 4 cos Θ
dx = -4 sin Θ dΘ</p>
<p>Substitute x, y, bounds, and dx all in terms of Θ back into the original integral.</p>
<p>∫xy dx [2,4] = ∫ 4cos Θ * 3sinΘ *-4sin Θ dΘ [pi/3, 0]
= -48 ∫ cos Θsin^2 Θ dΘ [pi/3, 0]</p>
<p>then you just use the fact that the negative of an integral is equal to the integral with the bounds reversed
so
= -48 ∫ cos Θsin^2 Θ dΘ [pi/3, 0]
= 48 ∫ cos Θsin^2 Θ dΘ [0, pi/3]</p>
<p>Edit: Yeah, the explanation in Barron’s is lacking a bit. Hopefully this helps.</p>
<p>Hm…tricky question. Thanks goodusername, that was very clear and concise :D:D Appreciate it.</p>