Thread: Evaluate the double integral Questions

1. Evaluate the double integral Questions

Question # 01:

Evaluate the double integral

Question # 02:

Prove that

Question # 03:
Use double integral to find the volume under the surface given by

over the region bounded by the curves

plz help me.

2. 1. $\int_{-1}^1{\int_{x^2}^1{(x^2 + y^2)\,dy}\,dx}$

$= \int_{-1}^1{\left[x^2y + \frac{1}{3}y^3\right]_{y=x^2}^{y =1}\,dx}$

$= \int_{-1}^1{\left[x^2 + \frac{1}{3}\right] - \left[x^4 + \frac{1}{3}x^6\right]\,dx}$

$= \int_{-1}^1{\frac{1}{3} + x^2 - x^4 - \frac{1}{3}x^6\,dx}$

$= \left[\frac{1}{3}x + \frac{1}{3}x^3 - \frac{1}{5}x^5 - \frac{1}{21}x^7\right]_{-1}^1$

$= \left[\frac{1}{3} + \frac{1}{3} - \frac{1}{5} - \frac{1}{21}\right] - \left[-\frac{1}{3} - \frac{1}{3} + \frac{1}{5} + \frac{1}{21}\right]$

$= \frac{4}{3} - \frac{2}{5} - \frac{2}{21}$.

I'll let you do the simplifying.

3. Originally Posted by sajjad002
Question # 02:

Prove that

2. is easy.

Since the limits on the integration are all constants it doesn't make any difference when you reverse the order of the integration.

4. thanks

5. So wat abt the third question ... any body could solve that one .. or just give some hints in that.. and in Q2 also!

6. Originally Posted by Angel Rox
So wat abt the third question ... any body could solve that one .. or just give some hints in that.. and in Q2 also!
I already GAVE you the answer to question 2.

If you're reversing the order of integration, you only need to change the terminals on the integrals when the terminals are functions of a variable.

Since they're constants, they stay the same.