Video Transcript
Evaluate the double integral, the
integral between 𝜋 by two and zero of the integral between 𝑦 and zero of cos 𝑥
sin 𝑦 d𝑥 d𝑦.
Double integrals are a way to
integrate over a two-dimensional area. Among other things, they let us
work out the volume under a surface. As the name suggests, it involves
integrating twice. In this case, our first step is to
integrate cos 𝑥 sin 𝑦 with respect to 𝑥 between the limit zero and 𝑦. We’ll then integrate this answer
with respect to 𝑦 between the limit zero and 𝜋 by two.
As stated, our first step is to
integrate cos 𝑥 sin 𝑦 with respect to 𝑥. As sin 𝑦 contains no 𝑥 term, this
can be treated as a constant. We, therefore, need to integrate
cos 𝑥 with respect to 𝑥 and multiply it by the constant sin 𝑦. The integral of cos 𝑥 d𝑥 is a
standard integral we need to remember. It is equal to the sin of 𝑥. This means that the integral of cos
𝑥 sin 𝑦 d𝑥 is sin 𝑥 sin 𝑦. As always with definite integrals,
we now need to substitute in the upper limit and the lower limit and subtract our
answers.
Substituting in the upper limit
gives us sin 𝑦 multiplied by sin 𝑦. And substituting in the lower limit
gives us sin of zero multiplied by sin of 𝑦. The sin of zero is equal to
zero. Therefore, the second term equals
zero. sin 𝑦 multiplied by sin 𝑦 can be rewritten as sin squared 𝑦. We have now completed the first
part of the integration. The integral between 𝑦 and zero of
cos 𝑥 sin 𝑦 d𝑥 is sin squared 𝑦.
We now need to integrate this
answer with respect to 𝑦 between zero and 𝜋 by two. One of our double angle formulae
states that cos of two 𝑥 is equal to one minus two sin squared 𝑥. This can be rearranged so that two
sin squared 𝑥 is equal to one minus cos two 𝑥. By dividing both sides of this
equation by two, sin squared 𝑥 is equal to one-half minus cos two 𝑥 over two. This means that we can rewrite the
integral of sin squared 𝑦 as the integral of a half minus cos two 𝑦 over two. We do this as there is no standard
integral for sin squared 𝑦.
Integrating a half with respect to
𝑦 gives us a half 𝑦 and integrating negative cos two 𝑦 over two gives us negative
sin of two 𝑦 over four. This is because the integral of cos
of 𝑛𝑥 is equal to sin of 𝑛𝑥 over 𝑛. We now need to substitute in the
limits 𝜋 over two and zero and subtract our two answers. Substituting in 𝜋 over two gives
us a half multiplied by 𝜋 over two minus sin of two multiplied by 𝜋 over two
divided by four. One-half multiplied by 𝜋 over two
is 𝜋 over four. sin of two multiplied by 𝜋 over two is the same as sin of 𝜋. The sin of 𝜋 is equal to zero. Therefore, the sin of 𝜋 divided by
four is also equal to zero.
Substituting in our upper limit of
𝜋 over two gives us 𝜋 over four minus zero. This is equal to 𝜋 over four. Substituting in our lower limit of
zero gives us a half multiplied by zero minus sin of two multiplied by zero divided
by four. One-half multiplied by zero equals
zero. Two multiplied by zero equals zero
and the sin of zero also equals zero. This means that our lower limit of
zero just gives us an answer of zero. 𝜋 by four minus zero equals 𝜋 by
four.
We can, therefore, say that the
final answer for the double integral is 𝜋 over four.
