How do you calculate the double integral of #(xcos(x+y))dr# where r is the region: 0 less than or equal to x less then or equal to #(2pi)/6#, 0 less than or equal to y less than or equal to #(2pi)/4#?

Answer 1
#int_R intxcos(x+y) dR#
Where #R = {(x, y) : 0 <= x <= pi/3, "and" 0 <= y <= pi/2}#

(I've reduced the fractions.)

We can choose whether to integrate first with respect to #x# or w.r.t. #y#. I'll find:
#int_0^(pi/3) int_0^(pi/2) xcos(x+y) dy dx = int_0^(pi/3) [int_0^(pi/2) xcos(x+y) dy] dx#
I'll do the inner integral, then substitute. #int_0^(pi/2) xcos(x+y) dy = xsin(x+y)]_(y=0)^(y = pi/2) = xsin(x+ pi/2) - xsin(x)#

Now we need to find

#int_0^(pi/3) [xsin(x+ pi/2) - xsin(x)] dx#

OK, maybe that wasn't easiest. Now I have to do 2 integration by parts, but rather than re-starting, we'll keep on this path.

We could replace #sin(x+ pi/2)# by #cos x#, but many students won't notice that and it's probably not any simpler, so I'll keep on with the problem as I have it above.

First Integral:

#int_0^(pi/3) x sin(x+ pi/2) dx#
Let #u=x# and #dv = sin(x+ pi/2) dx#
so #du = dx# and #v = -cos(x+ pi/2)#
#int_0^(pi/3) x sin(x+ pi/2) dx = -xcos(x + pi/2) + int cos(x + pi/2) dx#
# = -xcos(x + pi/2) + sin(x + pi/2)]_0^(pi/3)#
#= [- (pi)/3 cos((5 pi)/6) + sin((5 pi)/6)] - [0 + sin(pi/2)]#
#= - pi/3 (- sqrt3 / 2) +1/2 - 1 = (pi sqrt3)/6 -1/2 = (pi sqrt 3 -3)/6#

Second Integral:

The second integral: #int_0^(pi/3) [ - xsin(x)] dx# is almost the same as the first integral. Using the same method, we'll get
# = - (-xcos(x) + sin(x))]_0^(pi/3) = = xcos(x) - sin(x))]_0^(pi/3)#

Which evaluates to:

#(pi)/3 1 /2 - sqrt3/2 -[0cos (0) - sin(0)] = (pi - 3sqrt3)/6#

Adding the two integrals gives us:

#int_0^(pi/3) [xsin(x+ pi/2) - xsin(x)] dx = (pi sqrt 3 -3)/6+(pi - 3sqrt3)/6 = (pi sqrt3 - 3 sqrt3 + pi -3)/6 #
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

Could you clarify your question more? For one thing, you are sort of mixing rectangular and polar coordinates in the way you are stating it.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

To calculate the double integral of (x\cos(x+y)) over the region (r), where (0 \leq x \leq \frac{2\pi}{6}) and (0 \leq y \leq \frac{2\pi}{4}), you integrate with respect to (x) first and then with respect to (y).

The integral is evaluated as follows:

[ \int_{0}^{\frac{2\pi}{4}} \int_{0}^{\frac{2\pi}{6}} x\cos(x+y) ,dx,dy ]

First, integrate with respect to (x):

[ \int x\cos(x+y) ,dx = x\sin(x+y) - \int \sin(x+y) ,dx ]

[ = x\sin(x+y) + \cos(x+y) + C ]

Then evaluate the integral from (0) to (\frac{2\pi}{6}):

[ \left[x\sin(x+y) + \cos(x+y)\right]_{0}^{\frac{2\pi}{6}} ]

[ = \left[\left(\frac{2\pi}{6}\right)\sin\left(\frac{2\pi}{6}+y\right) + \cos\left(\frac{2\pi}{6}+y\right)\right] - \left[0\sin(0+y) + \cos(0+y)\right] ]

[ = \left(\frac{\pi}{3}\sin\left(\frac{\pi}{3}+y\right) + \cos\left(\frac{\pi}{3}+y\right)\right) - \cos(y) ]

Now, integrate the result with respect to (y) from (0) to (\frac{2\pi}{4}):

[ \int_{0}^{\frac{2\pi}{4}} \left(\frac{\pi}{3}\sin\left(\frac{\pi}{3}+y\right) + \cos\left(\frac{\pi}{3}+y\right) - \cos(y)\right) ,dy ]

[ = \left[-\frac{\pi}{3}\cos\left(\frac{\pi}{3}+y\right) + \sin\left(\frac{\pi}{3}+y\right) - \sin(y)\right]_{0}^{\frac{2\pi}{4}} ]

[ = \left(-\frac{\pi}{3}\cos\left(\frac{\pi}{3}+\frac{2\pi}{4}\right) + \sin\left(\frac{\pi}{3}+\frac{2\pi}{4}\right) - \sin\left(\frac{2\pi}{4}\right)\right) - \left(-\frac{\pi}{3}\cos\left(\frac{\pi}{3}\right) + \sin\left(\frac{\pi}{3}\right) - \sin(0)\right) ]

[ = \left(-\frac{\pi}{3}\cos\left(\frac{5\pi}{6}\right) + \sin\left(\frac{5\pi}{6}\right) - 0\right) - \left(-\frac{\pi}{3}\cos\left(\frac{\pi}{3}\right) + \sin\left(\frac{\pi}{3}\right) - 0\right) ]

[ = \left(-\frac{\pi}{3}\cdot\left(-\frac{\sqrt{3}}{2}\right) + \frac{1}{2} - 0\right) - \left(-\frac{\pi}{3}\cdot\left(\frac{1}{2}\right) + \frac{\sqrt{3}}{2} - 0\right) ]

[ = \frac{\pi\sqrt{3}}{6} + \frac{1}{2} - \frac{\pi}{6} - \frac{\sqrt{3}}{2} ]

[ = \frac{\pi\sqrt{3} + 3 - \pi\sqrt{3} - 3}{6} ]

[ = \frac{0}{6} ]

[ = 0 ]

Therefore, the double integral of (x\cos(x+y)) over the given region (r) is (0).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7