I am using xml in VBA.  The code below adds an item to a shopping cart on a remote website (posts a form).  Then the code displays the result in Internet Explorer.

You can see in the response there is a "get estimates" button.  I need to automatically click that, enter location info, and get the response of shipping and tax charges (for the item currently in the cart) on my excel 2010 worksheet.  

I want all the automated clicking and entering data to happen with the site (the site's server?) directly like when I added the item to the shopping cart, not through the browser if possible.  It takes a long time for the page to load and I think if I go through a browser I could do that without xml anyway.  But I'm really stuck so if I have to load a browser that's okay too.

Option Explicit

Sub testing()
    Dim objIE As Object
    Dim xmlhttp As Object
    Dim response As String
     Set objIE = CreateObject("InternetExplorer.Application")
    objIE.navigate "about:blank"
    objIE.Visible = True
    Set xmlhttp = CreateObject("MSXML2.ServerXMLHTTP")
    '~~&gt; Indicates that page that will receive the request and the type of request being submitted
    xmlhttp.Open "POST", "http://www.craft-e-corner.com/addtocart.aspx?returnurl=showproduct.aspx%3fProductID%3d2688%26SEName%3dnew-testament-cricut-cartridge", False
    '~~&gt; Indicate that the body of the request contains form data
    xmlhttp.setRequestHeader "Content-Type", "application/x-www-form-urlencoded"
    '~~&gt; Send the data as name/value pairs
    xmlhttp.Send "Quantity=1&amp;VariantID=2705&amp;ProductID=2688"
     response = xmlhttp.responseText
    objIE.Document.Write response
    Set xmlhttp = Nothing
End Sub



Question: How do I find the area between two polar curves?
Answer:  Integrate[Max[0, (4 + 4 Cos[t])^2/2 - 6^2/2 ], {t, -Pi, Pi}]


or 

 Integrate[(4 + 4 Cos[t])^2/2 - 6^2/2 , {t, -Pi/3, Pi/3}]



  18 Sqrt[3] - 4 Pi


Edit: another approach:

 Area[{r Sin[t], r Cos[t]}, {t, -Pi, Pi}, {r, 6, Max[6, 4 + 4 Cos[t]]}] 


or

 Area[CoordinateTransform[ "Polar" -&gt; "Cartesian", {r, t}],
                         {t, -Pi, Pi}, {r, 6, Max[6, 4 + 4 Cos[t]]}]



  2 (9 Sqrt[3] - 2 Pi)